Verification Of The Relation Between The Angles And Sides Of A Triangle

(i) The given triangle with naming is as shown below:

Given ∠ABC=35°, ∠BAC=45°

To find the exterior angle, i.e., ∠ACD=x°

Now Consider the ΔABC,

We know that the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in this case,

∠ACD=∠ABC+∠BAC

Substituting the values, we get

x°=35°+45°

⇒ x°=80°

So the measurement of exterior angle in this triangle is 80°.

(ii) The given triangle with naming is as shown below:

Given ∠ACB=35°, ∠BAC=50°

To find the exterior angle, i.e., ∠DBC=x°

Now Consider the ΔABC,

We know that the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in this case,

∠DBC=∠ACB+∠BAC

Substituting the values, we get

x°=50°+35°

⇒ x°=85°

So the measurement of exterior angle in this triangle is 85°.

(iii) The given triangle with naming is as shown below:

Given ∠ACB=50°, ∠BAC=40°

To find the exterior angle, i.e., ∠DBC=x°

Now Consider the ΔABC,

We know that the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in this case,

∠DBC=∠ACB+∠BAC

Substituting the values, we get

x°=50°+40°

⇒ x° =90°

So the measurement of exterior angle in this triangle is 90°.

We know in a triangle the sum of all three interior angles is equal to 180°.

So in ΔPRS,

∠PQR+∠QRP+∠RPQ=180°

⇒ ∠PQR+∠RPQ =180°-∠QRP………….(i)

Now QRS is a straight line, so

∠QRS=180°

⇒ ∠QRS=∠QRP+∠PRS=180°

⇒ ∠PRS=180°-∠QRP………(ii)

Substituting equation (ii) in equation (i), we get

∠PQR+∠RPQ=∠PRS

Hence the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

(i) The given triangle with naming is as shown below:

Given ∠ACB=55°, ∠BAC=40°

To find the angle, i.e., ∠ABC=x°

Now Consider the ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠ABC+∠BAC+∠ACB=180°

Substituting the values, we get

x°+40°+55°=180°

⇒ x°=180°-40°-55°

⇒ x°=85°

So the measurement of unknown angle in this triangle is 85°.

(ii) The given triangle with naming is as shown below:

Given ∠ACB=30°, ∠BAC=30°

To find the angle, i.e., ∠ABC=x°

Now Consider the ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠ABC+∠BAC+∠ACB=180°

Substituting the values, we get

x°+30°+30°=180°

⇒ x°=180°-30°-30°

⇒ x°=120°

So the measurement of unknown angle in this triangle is 120°.

(iii) The given triangle with naming is as shown below:

Given triangle is a right-angled triangle, so ∠ABC=90° and also given ∠ACB=30°

To find the angle, i.e., ∠BAC=x°

Now Consider the ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠ABC+∠BAC+∠ACB=180°

Substituting the values, we get

90°+x°+30°=180°

⇒ x°=180°-90°-30°

⇒ x°=60°

So the measurement of unknown angle in this triangle is 60°.

The given figure is as shown below:

Given ∠ABC = 40°, ∠BAD = 60°, ∠ADC = 20°

To find the exterior angle, i.e., ext ∠BCD = x°

Now the given figure is a concave quadrilateral; we will divide it into two by drawing a line joining AC.

By doing this we will get two triangles,

⇒ Quadrilateral ABCD = ΔABC + ΔADC

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

⇒ Quadrilateral ABCD = 180° + 180° = 360°

So in a quadrilateral the sum of all four interior angles is equal to 360°.

So in the given quadrilateral,

∠ABC + ∠BCD + ∠ADC + ∠BAD = 360°

Substituting the values, we get

40° + ∠BCD + 20° + 60° = 360°

⇒ ∠BCD = 360° - 60° - 20° - 40°

⇒ int. ∠BCD = 240°

From figure ∠BCD is reflex angle, so

360° = interior ∠BCD + Exterior ∠BCD

⇒Ext ∠ BCD = 360° - int. ∠BCD

Substituting the values, we get

⇒Ext ∠ BCD = 360° - 240°

⇒Ext ∠ BCD = 120°

So the measurement of exterior angle in the given figure is 120°.

The given figure is as shown below:

Given ∠PRQ = 60°, ∠QPR = 50°, ∠QST = 30°

To find the exterior angle, i.e., ext ∠STR = x°

Now consider ΔPQR

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠PRQ + ∠PQR + ∠QPR = 180°.

Substituting the values, we get

60° + ∠PQR + 50° = 180°

⇒ ∠ PQR = 180° - 60° - 50°

⇒ ∠ PQR = 70°

Now PQS is a straight line, so

∠PQS = 180°

⇒ ∠PQR + ∠SQT = 180°

Now substituting the values, we get

⇒ 70° + ∠SQT = 180°

⇒ ∠SQT = 180° - 70°

⇒ ∠SQT = 110°

From figure consider ΔQST, so

In this triangle ∠STR is exterior angle,

And we know the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

Hence ∠STR = ∠QST + ∠SQT

Substituting the values, we get

⇒∠STR = 30° + 110°

⇒∠STR = 140°

So the measurement of exterior angle in the given figure is 140°.

The given figure is as shown below:

Given ∠PQT = 55°, ∠RST = 60°

To find the angle, i.e., ∠TRS = x°

Now in the given figure PQ is parallel to TS, so

∠PQT = ∠RTS as they form is alternate interior angles.

So ∠RTS = 55°….(i)

Now consider ΔRTS,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠RTS + ∠TRS + ∠RST = 180°

Substituting the values from given criteria and equation(i), we get

55° + ∠TRS + 60° = 180°

⇒ ∠TRS = 180° - 60° - 55°

⇒ ∠TRS = 65°

So the measurement of unknown angle in the given figure is 65°.

Given: AB||CD, FE as the transversal line, ∠AFH = 110°, ∠HGE = 60°

To find the angles of ΔEHG, i.e., ∠GHE and ∠GEH

Now in the given figure, PQ is parallel to TS, so

∠AFH = ∠FHG as they form is alternate interior angles.

So ∠FHG = 110°….(i)

Now FHE is a straight line, so

∠FHE = 180°

⇒∠FHG + ∠GHE = 180°

Now substituting the value from equation (i), we get

⇒110° + ∠GHE = 180°

⇒∠GHE = 180° - 110°

⇒∠GHE = 70°……..(ii)

Now consider ΔEHG,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠GHE + ∠GEH + ∠HGE = 180°

Substituting the values from given criteria and equation(ii), we get

70° + ∠GEH + 60° = 180°

⇒ ∠GEH = 180° - 60° - 70°

⇒ ∠GEH = 50°

So the measurement of the unknown angle of the given triangle EHG are 50° and 70°.

Given: from the given figure AB||CF

To find the measurement of ∠A + ∠B + ∠C + ∠D + ∠E + ∠F

Now in the given figure, AB is parallel to CF with AD as tranversal line, so

∠A = ∠COD……….(i) (as they form is corresponding angles)

Similarly, AB is parallel to CF with BE as tranversal line, so

∠B = ∠FOE……….(ii) (as they form is corresponding angles)

Now consider ΔCOD,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠C + ∠D + ∠COD = 180°

Substituting the values from equation(i), we get

∠C + ∠D + ∠A = 180°……..(iii)

Now consider ΔFOE,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠E + ∠F + ∠FOE = 180°

Substituting the values from equation(ii), we get

∠E + ∠F + ∠B = 180°……..(iv)

Now adding equation (iii) and equation (iv), we get

∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 180° + 180° = 360°

So the measurement of ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°.

Given ΔABC, ∠ACD = 112° and AB = AC

To find the measurement of ∠ABC, ∠ACB and ∠BAC

In the given figure, BCD is a straight line, so

∠BCD = 180°

But, ∠BCD = ∠ACB + ∠ACD

∴ ∠ACB + ∠ACD = 180°

But given ∠ACD = 112°, so the above equation becomes,

⇒ ∠ACB + 112° = 180°

⇒ ∠ACB = 180° - 112°

⇒ ∠ACB = 68°………(i)

Now consider ΔABC, it is given AB = AC

We know angles opposite to equal sides are equal, therefore

⇒∠ACB = ∠ABC

Now equation (i) in above equation we get

∠ABC = 68°……….(ii)

Considering the same triangle, i.e., ΔABC

We know in a triangle the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in this case ∠ACD is an exterior angle, therefore

∠ACD = ∠ABC + ∠BAC

Substituting the values from given criteria and equation (ii), we get

112° = 68° + ∠BAC

⇒ ∠BAC = 112° - 68°

⇒ ∠BAC = 44°……..(iii)

Hence from equation (i), (ii) and (iii), the measurements are

∠ABC = 68°, ∠ACB = 68° and ∠BAC = 44°

Given: ΔABC, ∠BAC = 80° and AB = AC

To find the measurement of ∠ABC and ∠ACB

Now in ΔABC, it is given AB = AC

We know angles opposite to equal sides are equal, therefore

⇒∠ACB = ∠ABC………(i)

We also know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ACB + ∠ABC = 180°

Substituting the values from equation(i) and given criteria, we get

80° + ∠ABC + ∠ABC = 180°

⇒ 80° + 2∠ABC = 180°

⇒ 2∠ABC = 180° - 80°

⇒ 2∠ABC = 100°

⇒ ∠ABC = 50°

Substituting the above value in equation (i) we get

⇒∠ACB = ∠ABC = 50°

These are the required values of the angles.

Given: ΔABC, ∠ABC = 70° and AB = AC

To find the measurement of ∠ACB and ∠BAC

Now in ΔABC, it is given AB = AC

We know angles opposite to equal sides are equal, therefore

⇒∠ACB = ∠ABC

Substituting given value in the above equation, we get

⇒∠ACB = 70°…….(i)

We also know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ACB + ∠ABC = 180°

Substituting the values from equation(i) and given criteria, we get

∠BAC + 70° + 70° = 180°

⇒ ∠BAC = 180° - 70° - 70°

⇒ ∠BAC = 40°

From equation (i) and (ii), the required values of the angles are

∠ACB = 70° and ∠BAC = 40°

Given: ΔABC, ∠BAC + ∠ACB = 50° and AB = BC

To find the measurement of angles of the triangle ABC

Now in ΔABC, it is given AB = BC

We know angles opposite to equal sides are equal, therefore

⇒∠BAC = ∠ACB……….(i)

Given ∠BAC + ∠ACB = 50°

Substituting values from equation (i) in the above equation, we get

⇒∠ACB + ∠ACB = 50°

⇒2∠ACB = 50°

⇒∠ACB = 25°…….(ii)

Substituting values from equation (ii) in equation (i), we get

⇒∠BAC = ∠ACB = 25°……….(iii)

We also know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ACB + ∠ABC = 180°

Substituting the values from equation(iii) and given criteria, we get

∠ABC + 25° + 25° = 180°

⇒ ∠ABC = 180° - 25° - 25°

⇒ ∠ABC = 130°……(iv)

From equation (iii) and (iv), the required values of the angles are

∠BAC = ∠ACB = 25°, ∠ABC = 130°

Given: ΔABC, O is an interior point of ΔABC

To prove that ∠BOC > ∠BAC

The figure for the given question is as shown below,

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ACB + ∠ABC = 180°………(i)

But from the above figure,

∠ABC = ∠ABO + ∠OBC………(ii)

And also,

∠ACB = ∠ACO + ∠OCB…….(iii)

Substituting equation (ii) and (iii) in equation (i), we get

∠BAC + (∠ACO + ∠OCB ) + ( ∠ABO + ∠OBC ) = 180°

⇒ ∠OBC + ∠OCB = 180° - ∠BAC - ∠ACO - ∠ABO………(iv)

Now consider ΔBOC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BOC + ∠OCB + ∠OBC = 180°

Substituting equation (iv) in the above equation, we get

∠BOC + (180° - ∠BAC - ∠ACO - ∠ABO ) = 180°

⇒ ∠BOC = 180° - (180° - ∠BAC - ∠ACO - ∠ABO )

⇒ ∠BOC = ∠BAC + ∠ACO + ∠ABO

⇒ ∠BOC > ∠BAC

Hence proved

Given: ΔABC, exterior angles ∠ABD and ∠ACE

To prove: the sum of the measurement of these two exterior angles is more than 2 right angles, i.e., ∠ABD + ∠ACE > 2(90°)

The figure for the given question is as shown below,

We know in a triangle the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in ΔABC ∠ABD and ∠ACE are exterior angles, so

∠ABD = ∠BAC + ∠ACB……..(i)

And,

∠ACE = ∠ABC + ∠BAC……..(ii)

Adding equation (i) an equation (ii), we get

∠ABD + ∠ACE = ∠BAC + ∠ACB + ∠ABC + ∠BAC …….(iii)

We also know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ACB + ∠ABC = 180°

Substituting the above equation in equation (iii), we get

∠ABD + ∠ACE = (∠BAC + ∠ACB + ∠ABC) + ∠BAC

⇒ ∠ABD + ∠ACE = (180°) + ∠BAC

⇒ ∠ABD + ∠ACE > 180°

⇒ ∠ABD + ∠ACE > 2(90°)

Hence the sum of the measurement of these two exterior angles is more than 2 right angles.

Hence proved

Given: ΔABC, line l||BC, line m||BA, line l and line m are meeting at point D

To prove: ∠ABC = ∠ADC

The figure for the given question is as shown below,

Now in the given figure BA is parallel to line m with AC as tranversal line, so

∠BAC = ∠ACD……….(i) (as they form is alternate interior angles)

Similarly, BC is parallel to line l with AC as tranversal line, so

∠ACB = ∠CAD……….(ii) (as they form is alternate interior angles)

Now consider ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠ABC + ∠BAC + ∠ACB = 180°

Substituting the values from equation(i) and (ii), we get

∠ABC + ∠ACD + ∠CAD = 180°

⇒ ∠ABC = 180° - ∠ACD - ∠CAD…………(iii)

Now consider ΔADC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠ADC + ∠CAD + ∠ACD = 180°

⇒ ∠ADC = 180° - ∠CAD - ∠ACD…..(iv)

Equating equation (iii) and equation (iv), we get

∠ABC = ∠ADC

Hence proved

Given: ΔABC, internal angle bisectors of ∠ABC and ∠ACB meet at O

To prove:

Let BD be the internal angle bisector of ∠ABC and CE be the internal angle bisector of ∠ACB. The figure of the above question is as shown below,

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ACB + ∠ABC = 180°………(i)

But from above figure,

∠ABC = ∠ABO + ∠OBC……(ii)

And it is also given BD is internal angle bisector of ∠ABC, so

∠ABO = ∠OBC

Substituting the above value in equation (ii), we get

⇒ ∠ABC = 2∠OBC………(iii)

And also from figure,

∠ACB = ∠ACO + ∠OCB……….(iv)

And it is also given CE is internal angle bisector of ∠ACB, so

∠ACO = ∠OCB

Substituting the above value in equation (iv), we get

⇒ ∠ACB = 2∠OCB………(v)

Substituting equation (iii) and (v) in equation (i), we get

∠BAC + (2∠OCB ) + ( 2∠OBC ) = 180°

⇒ 2(∠OBC + ∠OCB) = 180° - ∠BAC

Now consider ΔBOC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BOC + ∠OCB + ∠OBC = 180°

Substituting equation (vi) in above equation, we get

Hence proved

Given: ΔABC, external angle bisectors of ∠ABC and ∠ACB meet at O

To prove:

Let BO be the external angle bisector of ∠ABC, i.e., BO is angle bisector of ∠CBD, and CO be the external angle bisector of ∠ACB, i.e., CO is the angle bisector of ∠BCE. The figure of the above question is as shown below,

Now in ΔABC,

We know in a triangle the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in this case,

∠CBD = ∠ACB + ∠BAC………(i)

And also,

∠BCE = ∠ABC + ∠BAC……..(ii)

But it is also given BO is angle bisector of ∠CBD, so

∠DBO = ∠CBO

⇒ ∠CBD = 2∠CBO

Substituting the above value in equation (i), we get

2∠CBO = ∠ACB + ∠BAC

And it is also given CO is angle bisector of ∠BCE, so

∠ECO = ∠BCO

⇒ ∠BCE = 2∠BCO

Substituting the above value in equation (ii), we get

2∠BCO = ∠ABC + ∠BAC

Now consider ΔBOC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BOC + ∠BCO + ∠CBO = 180°

Substituting equation (iii) and (iv) in above equation, we get

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ACB + ∠ABC = 180°

Substituting the above value in equation (v), we get

Hence proved

Given: ΔABC, external angle bisectors of ∠ACB, line parallel to BC passing through point A meets

To prove:

Let CD be the external angle bisector of ∠ACB, i.e., CD is angle bisector of ∠ACE. And let line l be the parallel line to side BC through point A. So line l and CD meets at point D. The figure of the above question is as shown below,

Now in ΔABC,

We know in a triangle the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in this case,

∠ACE = ∠ABC + ∠BAC………(i)

But it is also given CD is angle bisector of ∠ACE, so

∠ACD = ∠DCE

⇒ ∠ACE = 2∠ACD

Substituting the above value in equation (i), we get

2∠ACD = ∠ABC + ∠BAC……..(ii)

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ABC + ∠ACB = 180°

Substituting the value from equation (ii) in above equation, we get

2∠ACD + ∠ACB = 180°

⇒ 2∠ACD = 180° - ∠ACB

Hence proved

Given: a triangle with vertical angle bisector, perpendicular from vertex to the base

To prove: the angle between the bisector of the vertical angle of a triangle and the perpendicular drawn from the vertex to the base of the triangle is half of the difference of the base angles

Let the triangle be ΔABC,

Let AE be the bisector of vertical angle A,

And Let AD be the perpendicular drawn from the vertex A to the base BC of the triangle ABC.

So the corresponding figure will be as shown below,

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ABC + ∠ACB = 180°………(i)

Given AE is angle bisector of ∠BAC

⇒ ∠BAE = ∠CAE

⇒ ∠BAC = 2∠BAE

Substituting the above value in equation (i) we get

2∠BAE + ∠ABC + ∠ACB = 180°………(ii)

But from figure, ∠BAE = ∠BAD + ∠DAE

Substituting this value in equation (ii), we get

2(∠BAD + ∠DAE) + ∠ABC + ∠ACB = 180°

⇒ 2∠BAD + 2∠DAE + ∠ABC + ∠ACB = 180°………(iii)

Given AD is perpendicular to BC, so ΔBAD and ΔDAE is right - angled triangle, hence

In right - angled ΔBAD,

∠ABD + ∠BAD = 90°

⇒ ∠ABC + ∠BAD = 90°

⇒ ∠BAD = 90° - ∠ABC…….(iv)

Substituting equation (iv) in equation (iii), we get

⇒ 2(90° - ∠ABC) + 2∠DAE + ∠ABC + ∠ACB = 180°

⇒ 180° - 2∠ABC + 2∠DAE + ∠ABC + ∠ACB = 180°

⇒ 180° - ∠ABC + 2∠DAE + ∠ACB = 180°

⇒ 2∠DAE = 180° - 180° + ∠ABC - ∠ACB

⇒ 2∠DAE = ∠ABC - ∠ACB

…..(v)

Here ∠ABC and ∠ACB are base angles and ∠DAE is the angle between the bisector of vertical angle and perpendicular from vertex to the base.

Hence the angle between the bisector of the vertical angle of a triangle and the perpendicular drawn from the vertex to the base of the triangle is half of the difference of the base angles.

Hence Proved

Given: isosceles ΔABC and one base angle is twice of the vertical angle

To find the measurement of three of the triangle

The figure for the above question is as shown below,

Now given ΔABC is an isosceles triangle,

AB = AC and ∠ABC = ∠ACB……..(i)

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ABC + ∠ACB = 180°

Substituting equation (i) in above equation we get

∠BAC + ∠ABC + ∠ABC = 180°

∠BAC + 2∠ABC = 180°………..(ii)

In this triangle ∠ ABC and ∠ACB are base angles

It is given base angle is twice of the vertical angle

⇒ ∠ABC = 2∠BAC…….(iii)

Substituting equation (iii) in equation (ii) we get

∠BAC + 2(2∠BAC) = 180°

⇒ ∠BAC + 4∠BAC = 180°

⇒ 5∠BAC = 180°

⇒ ∠BAC = 36°……(iv)

Substituting equation (iv) in equation (iii) we get

⇒ ∠ABC = 2(36°)

⇒ ∠ABC = 72°…….(v)

Substituting equation (v) in equation (i) we get

∠ACB = 72°…..(vi)

So from equation (iv), (v) and (vi), the measurements of the angles

of ΔABC are

∠BAC = 36°, ∠ABC = 72°, ∠ACB = 72°

To prove:

The figure for the above question is as shown below,

Given ΔABC is a right - angled triangle, now applying the properties of the triangle,

We know

Now when θ = 30°, AB is the opposite side and BC is the hypotenuse, then the above equation becomes,

Now we know the value of , substituting this in above equation, we get

Hence proved

Given: ΔXYZ, ∠XYZ = 90° and

To prove: ∠YXZ = 60°

The figure for the above question is as shown below,

Given ΔXYZ is a right - angled triangle, now applying the properties of the triangle,

We know

Now when θ = ∠XZY, XY is the opposite side and XZ is the hypotenuse, then the above equation becomes,

Given

Equating equation (i) and (ii), we get

But,

Hence θ = 30°

Or ∠XZY = 30°….(iii)

Now in ΔXYZ,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠XYZ + ∠XZY + ∠YXZ = 180°

Substituting given value and value from equation (iii) in above equation we get

90° + 30° + ∠YXZ = 180°

⇒ ∠YXZ = 180° - 90° - 30°

⇒ ∠YXZ = 60°

Hence proved

Given: an equilateral triangle

To prove the measurement of each angle of the equilateral triangle is 60°

Let ΔABC be an equilateral triangle,

We know all sides of the equilateral triangle are equal hence

Hence, AB = BC = AC

So we have to prove, ∠A = ∠B = ∠C = 60°

Now AB = AC

And we know angles opposite to equal sides are equal, so

⇒ ∠C = ∠B……..(i)

Also, AC = BC

And we know angles opposite to equal sides are equal, so

⇒ ∠B = ∠A……..(ii)

So from (i) and (ii), we get

∠A = ∠B = ∠C……….(iii)

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠A + ∠B + ∠C = 180°

Substituting value from equation (iii) in above equation we get

∠A + ∠A + ∠A = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

Substituting this back in equation (iii), we get

∠A = ∠B = ∠C = 60°

Hence the measurement of each angle of the equilateral triangle is 60°

Hence proved

Given: ΔABC, the bisector of ∠BAC, D is the midpoint of side AC, line parallel to side AB through point D, the bisector and parallel line meet at point E outside BC

To prove ∠AEC = 1 right angle, i.e., ∠AEC = 90°

In ΔABC, let AE be the bisector of ∠BAC,

Let DE be the parallel line to side AB though point D, where D is the midpoint of AC.

So the figure of the given question is as shown below,

Given AE is the bisector of ∠BAC, so let

∠BAE = ∠EAC = x……….(i)

Given DE||AB and with AE as the transversal line,

Then ∠BAE = ∠AED = x ……..(ii) (as they form alternate interior angles)

Now consider ΔADE,

From equation (ii),

∠BAE = ∠AED = x

We know sides opposite to equal angles are equal, so

AD = DE………(iii)

And in same triangle ∠EDC form external angle, but we know the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles, so

∠EDC = ∠EAD + ∠AED

Now substituting values from equation (i) and (ii), we get

∠EDC = x + x = 2x….(iv)

Now consider ΔAEC, given D is midpoint of AC,

So AD = DC……(iv)

We know angles opposite to equal sides are equal, so

∠AED = ∠DEC

Now substituting the value from equation (ii), we get

∠AED = DEC = x………(v)

By comparing equation (iii) and (iv), we get

DE = DC…..(v)

Now consider ΔDEC

We also know angles opposite to equal sides are equal, so from equation (v), we get

∠DEC = ∠ECD

Now substituting the value from equation (ii), we get

∠DEC = ∠ECD = x……….(vi)

Now in ΔEDC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠EDC + ∠DEC + ∠ECD = 180°

Substituting value from equation (iv) and (vi) in above equation we get

2x + x + x = 180°

⇒ 4x = 180°

⇒ x = 45°……….(vii)

Now from figure,

∠AEC = ∠AED + ∠DEC

Substituting values from equation (v), we get

∠AEC = x + x = 2x

Substituting values from equation (vii), we get

∠AEC = 2(45°) = 90°

Hence ∠AEC = 1 right angle

Hence proved

(i) The given triangle is as shown below:

Given PQ=2.5cm, QR=3.5cm, PR=4cm

To find the ∠P is greater than which angle.

In the given ΔPQR,

∠P is opposite to side QR,

And from the given values, side QR=3.5cm is greater than side PQ=2.5cm

Now from the given figure, ∠R is opposite to the side PQ.

⇒ QR>PQ

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ ∠P>∠R

Hence ∠P is greater than ∠R.

(ii) The given triangle is as shown below:

Given XY=4cm, YZ=5cm, XZ=6cm

To find the ∠x is greater than which angle.

In the given ΔXYZ,

∠X is opposite to side YZ,

And from the given values, side YZ=5cm is greater than side XY=4cm

Now from the given figure, ∠Z is opposite to the side XY.

⇒ YZ>XY

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ ∠X>∠Z

Hence ∠X is greater than ∠Z.

(iii) The given triangle is as shown below:

Given AB=4cm, BC=5cm, CA=3cm

To find the ∠C is greater than which angle.

In the given ΔABC,

∠C is opposite to side AB,

And from the given values, side AB=4cm is greater than side AC=3cm

Now from the given figure, ∠B is opposite to the side AC.

⇒ AB>AC

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ ∠C>∠B

Hence ∠C is greater than ∠B.

Now we will measure the length of the sides from each triangle using a scale, we get

First, consider ΔMAN

We know angles opposite to equal sides are equal, hence in ΔMAN,

Side AM= side MN,

⇒ ∠N=∠A

Hence in this ΔMAN, two angles are equal, so this is not what Pallabi or Siraj drew.

Now we will consider the next triangle, i.e., ΔPAN

In this, no two sides are equal, i.e.,

AN>PN

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

In ΔPAN, ∠P is opposite to side AN, and ∠A is opposite to side PN,

⇒ ∠P>∠A

Hence in these two angles are unequal, so this triangle is drawn either by Pallabi or Siraj.

Now we will consider the next triangle, i.e., ΔFAN

In this, no two sides are equal, i.e.,

FN>AF

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

In ΔFAN, ∠A is opposite to side FN, and ∠N is opposite to side AF,

⇒ ∠A>∠N

Hence in this two angles are unequal, so this triangle is drawn either by Pallabi or Siraj.

(i) The given triangle is as shown below:

Given ∠A=60°, ∠C=30°, and the given triangle is a right-angled triangle, so ∠B=90°

To find the side BC is greater of side AB is greater.

In the given ΔABC,

∠A is opposite to side BC,

And ∠C is opposite to side AB

And from the given values, ∠A=60°is greater than ∠C=30°,

⇒ ∠A>∠C

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ side BC>side AB

Hence side BC is greater than side AB.

(ii) The given triangle is as shown below:

Given ∠X=60°, ∠Y=75°, ∠Z=45°

To find the side YZ is greater than which side.

In the given ΔXYZ,

∠X is opposite to side YZ,

And from the given values, ∠X=60°is greater than ∠Z=45°,

And ∠Z is opposite to side XY

⇒ ∠X>∠Z

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ side YZ>side XY

Hence side YZ is greater than side XY.

(iii) The given triangle is as shown below:

Given ∠Q=59°, ∠R=50°

To find which side is greater than which side

In the given ΔPQR

⇒ ∠Q>∠R

And from the given figure,

∠Q is opposite to side PR,

And ∠R is opposite to side PQ

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ side PR>side PQ

Hence side PR is greater than side PQ.

Given: ΔPQR and ∠QPR > ∠PQR

To find the relation between PR and QR

In the given ΔPQR,

∠QPR is opposite to side QR,

And also ∠PQR is opposite to the side PR.

And given ∠QPR > ∠PQR

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ QR > PR

Hence this is the relation between QR and PR.

Given: ΔABC, AC > AB, D is point on the side AC such that ∠ADB = ∠ABD

To prove: ∠ABC > ∠ACB

The figure for the given question is as shown below,

In the given ΔABC,

∠ABC is opposite to side AC,

And also ∠ACB is opposite to the side AB.

And given AC > AB

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ ∠ABC > ∠ACB

Hence proved

Given, AD is the bisector of ∠A of triangle ABC .

Hence, ∠DAB = ∠DAC

Now, it is given that, AB > AC

Therefore,

∠BDA > ∠ ADC

Hence, BD < CD

Proved.

Given: ΔABC, AD is perpendicular to BC and AC > AB

To prove: (i) ∠CAD > ∠BAD

(ii) DC > BD

The figure to the given question is as shown below,

(i) Given AD is perpendicular to BC, so the ΔABC is divided into two right - angled triangles namely ΔABD and ΔADC.

Now consider ΔABD,

By Pythagoras theorem, we have

AB² = AD² + BD²……….(i)

Similarly in ΔADC,

by Pythagoras theorem, we have

AC² = AD² + DC²………(ii)

And it is given,

AC > AB

Squaring on both sides we get

AC² > AB²

Now substituting the values from equation (i) and (ii) in above equation, we get

AD² + DC² > AD² + BD²

Now cancelling the like terms on both sides we get

DC² > BD²

Taking square root on both sides, we get

DC > BD………(iii)

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒∠CAD > ∠BAD

Hence proved

(ii) From equation (iii),

DC > BD

Hence proved

Given: a quadrilateral such that the greatest and the smallest sides are opposite to each other

To prove: the adjacent angle to the greatest side is smaller than its opposite side

Let ABCD be the quadrilateral, then according to the given criteria the figure is as shown below,

Let us draw two diagonals BD and AC as shown in the figure.
In ΔABD,

AB < AD < BD (as AB is the smallest side of the quadrilateral)

So, ∠ADB < ∠ABD - - - - - - - - (1)
(angle opposite to smaller side is smaller)
In ΔBCD

BC < DC < BD (As CD is the longest side in the quadrilateral, given)
So, ∠BDC < ∠CBD - - - - - - - - - (2)

Adding equations (1) and (2)
∠ADB + ∠BDC < ∠ABD + ∠CBD
Or, ∠ADC < ∠ABC………..(i)

Similarly, in ΔABC

AB < BC < AC
∠ACB < ∠BAC - - - - - - - - - (3)
in ΔADC

AD < AC
∠DCA < ∠DAC - - - - - - - - (4)

Adding equations (3) and (4)
∠ACB + ∠DCA < ∠DAC + ∠BAC
Or, ∠BCD < ∠BAD…….(ii)

Hence from equation (i) and (ii), we can conclude that the adjacent angle to the greatest side is smaller than its opposite side.

Given: a figure such that AB < OB and CD > OD

To prove: ∠BAO > ∠OCD

Consider ΔABO,

Given AB < OB

From figure, AB is opposite to ∠AOB, and OB is opposite to ∠OAB

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒∠AOB < ∠OAB…………(i)

Now consider ΔCOD,

Given CD > OD

From figure, CD is opposite to ∠COD, and OD is opposite to ∠OCD

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒∠COD > ∠OCD…………(ii)

From figure we can see that

∠AOB = ∠COD……..(iii) (as they form vertically opposite angles)

Now substituting equation (iii) in equation (i), we get

⇒∠COD < ∠OAB…………(iv)

Now comparing equation (ii) and equation (iv), we get

∠OAB > ∠OCD

Or ∠BAO > ∠OCD

Hence Proved

Given: ΔPQR, PQ > PR, PS = PR

To prove: (i)

(ii)

The figure to the given question is as shown below,

(i) Now Consider the ΔPSR,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠PSR + ∠PRS + ∠P = 180°……(i)

But given PS = PR

And we know angles opposite to equal sides are equal, so

∠PSR = ∠PRS………(ii)

Substituting this value in equation (i), we get

∠PSR + ∠PSR + ∠P = 180°

⇒ 2∠PSR + ∠P = 180°

⇒ ∠P = 180° - 2∠PSR……(iii)

Now Consider the ΔPQR,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠PQR + ∠PRQ + ∠P = 180°

Substituting the value of ∠P from equation (iii), we get

∠PQR + ∠PRQ + 180° - 2∠PSR = 180°

⇒ 2∠PSR = ∠PQR + ∠PRQ + 180° - 180°

⇒ 2∠PSR = ∠PQR + ∠PRQ

(iv)

Hence proved

(ii) From equation (ii), ∠PSR = ∠PRS

And from equation (iv),

Comparing these two we get

From figure, ∠PRS = ∠PRQ - ∠QRS

Substituting this value in equation (v), we get

Hence Proved

Given: ΔABC, AB > AC, the bisector of ∠BAC meets BC at D, AE = AC

To prove: (i) ΔACD≅ΔAED

(ii) ∠ACB > ∠ABC.

The diagram for the given criteria is as shown below,

(i) Consider ΔAED and ΔACD

AD = AD (common)

∠EAD = ∠CAD (as AD is the bisector of ∠BAC)

AE = AC (given)

Hence by SAS congruency criteria,

ΔAED≅ΔACD

Hence proved

(ii) In the given ΔABC,

∠ACB is opposite to side AB

And ∠ABC is opposite to the side AC

And given AB > AC

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ ∠ACB > ∠ABC

Hence Proved

Given: a figure such that AB = CD, ∠OCD > ∠COD and ∠OAB < ∠AOB

To prove: OB < OD

Consider ΔABO,

Given ∠AOB > ∠OAB

From the figure, AB is opposite to ∠AOB, and OB is opposite to ∠OAB

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ AB > OB…………(i)

Now consider ΔCOD,

Given ∠OCD > ∠COD

From figure, CD is opposite to ∠COD, and OD is opposite to ∠OCD

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒OD > CD…………(ii)

And given AB = CD………..(iii)

Now substituting equation (iii) in equation (i), we get

⇒ CD > OB…………(iv)

Now comparing equation (ii) and equation (iv), we get

OD > OB

Or OB < OD

Hence Proved.

Given: a right - angled triangle

To prove: the hypotenuse is the greatest side

Let ΔABC be the right angled triangle, right angled at B as shown below,

Now Consider the ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠ABC + ∠BAC + ∠ACB = 180°

But the given triangle is right angled, right angle at ∠ABC

So when ∠ABC = 90°, then the other two angles of the triangle together will be equal to 90°, as the sum of all three angles of a triangle is equal to 180°.

Hence ∠ABC > ∠BAC or ∠ABC > ∠ACB

But in a triangle we know the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle

Hence AC > BC or AC > AB

Therefore the hypotenuse of a right angled triangle is the greatest side.

Hence proved

Given: an obtuse angled triangle

To prove: the measurement of the opposite side is the greatest side

Let ΔABC be the obtuse angled triangle as shown below,

Now Consider the ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠ABC + ∠BAC + ∠ACB = 180°

But the given triangle is obtuse angled, obtuse at ∠ABC

But we know obtuse angle means angle is greater than 90°.

So when ∠ABC is greater than 90°, the other two angles of the triangle should be less than 90°, as the sum of all three angles of a triangle is equal to 180°.

Hence ∠ABC > ∠BAC or ∠ABC > ∠ACB

But in a triangle we know the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle

Hence AC > BC or AC > AB

Therefore the measurement of the opposite side of the obtuse angle is the greatest side in obtuse angled triangle.

Given: ΔABC, the bisectors of ∠ABC and ∠ACB meet at I and AB > AC

To prove: IB > IC

The figure for the given criteria is as shown below,

Now given BD is bisector of ∠ABC, so

∠ABD = ∠DBC

Or ∠ABC = 2∠DBC…………(i)

Now given CE is bisector of ∠ACB, so

∠ACE = ∠ECB

Or ∠ACB = 2∠ECB…………(ii)

Given AB > AC

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ ∠ACB > ∠ABC

Now substituting values from equation (i) and (ii) in above equation, we get

⇒ 2∠ECB > 2∠DBC

⇒ ∠ECB > ∠DBC…….(iii)

Now consider ΔIBC,

From equation (iii), we have

∠ICB > ∠IBC

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

Hence IB > IC

Hence proved