Lines And Angles

From the given figure, we have

AB || CD || EF

PQ || RS

∠RQD = 25°

∠CQP = 60°

Now, we have PQ || RS.

We know if a transversal intersects two parallel lines, then each pair of alternate exterior angles is equal.

Since, PQ || RS

⇒ ∠PQC = ∠BRS

We have ∠PQC = 60°

⇒ ∠BRS = 60° - - - - (i)

We know, if a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Since, AB || CD

⇒ ∠DQR = ∠QRA

We have ∠DQR = 25°

⇒ ∠QRA = 25° - - - - (ii)

By linear pair axiom,

∠ARS + ∠BRS = 180°

⇒ ∠ARS = 180° - ∠BRS

⇒ ∠ARS = 180° - 60° (From (i), ∠BRS = 60°)

⇒ ∠ARS = 120° - - - - (iii)

Now, ∠QRS = ∠QRA + ∠ARS

From equations (ii) and (iii), we have ∠QRA = 25° and ∠ARS = 120°

Thus, the above equation can be written as:

∠QRS = 25° + 120°

⇒ ∠QRS = 145°

Thus, the option (C) is correct.

Option (A) is not correct because 85°, i.e., 60° + 25° is not the correct angle combination for ∠QRS.

Option (B) is not correct because 135°, is not the exact solution of ∠QRS.

Option (D) is not correct because 110°, is not the correct solution of ∠QRS.

Let us draw a triangle ABC.

We know, sum of all the angles of a triangle is 180°.

So, ∠A + ∠B + ∠C = 180° - - - - (i)

It is also mentioned in the question that one angle of the triangle is equal to the sum of the other two triangle.

Let us assume that, ∠A + ∠B = ∠C - - - - (ii)

From equation (i), we can write,

∠A + ∠B + ∠C = 180°

From equation (ii), the above equation can be written as

∠A + ∠B + ∠A + ∠B = 180°

⇒ 2 × (∠A + ∠B) = 180°

⇒ ∠A + ∠B = 90° - - - - (iii)

From equation (ii), we have

∠A + ∠B = ∠C

So, from equation (iii), we get

∠A + ∠B = ∠C = 90°

From above, it is clear that one angle of the triangle, here ∠C, is a right angle, since its value is 90°, and this is equal to the sum of the other two angles.

Therefore, this type of triangle whose one angle is equal to the sum of the other two angles is a right triangle.

Thus, the option (D) is correct.

Option (A) is not correct because in an isosceles triangle at least two sides of the triangle are equal. This implies that the two opposite angles should also be equal. It is given that one angle of the triangle, i.e., ∠C is equal to the sum of the other two angles, i.e., ∠A and ∠B. Only one case exists where ∠A = ∠B = 45°, such that ∠A + ∠B = ∠C = 90°. In all other values for ∠A, ∠B, and ∠C, the equation doesn’t hold true. So, it is not an isosceles triangle.

Option (B) is not correct because in an obtuse triangle, one of the angles is an obtuse angle, i.e., greater than 90°. But, we find out from the result that one of the angle, i.e., ∠C is 90° and the other two angles, i.e., ∠A and ∠B are the sum of the third angle, ∠C which is a right angle. Since, the sum of the two angles is equal to the third angle and the third angle is 90°, therefore neither the first nor the second angle can be greater than 90°. As none of the three angles of the triangle is more than 90°, it is not an obtuse triangle.

Option (C) is not correct because in an equilateral triangle, all the three sides of the triangle are equal. This implies, in an equilateral triangle, all the three angles are equal. But it is given to us that one angle of the triangle is equal to the sum of the other two angles where we have found one angle, i.e., ∠C as right angle. We know that sum of the angles of a triangle is equal to 180°. Since, one of the angle of the triangle, ∠C is 90°, this means that the none of the other two triangles is equal to 90°. This is in accordance with the linear pair axiom. As the three angles of the triangle are unequal, it is not an equilateral triangle.

The diagram for the question is as -

Since QR is extended to a point S, ∠PRS becomes the exterior angle of the ΔPQR.

It is given to us that an exterior angle of a triangle is 105°.

So, ∠PRS = 105° - - - - (i)

It is known to us that -

If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

The two interior opposite angles are ∠RPQ and ∠PQR.

So, - - - - (ii)

But, in the question it is given that the interior opposite angles are equal.

⇒ ∠RPQ = ∠PQR - - - - (iii)

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

⇒

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

⇒ ∠RPQ =

From (iii), we have ∠RPQ = ∠PQR

⇒

Thus, each of these equal angles is . So, option (B) is correct.

Also, (Since, sum of all the angles of a triangle is 180°)

And, (By linear pair axiom)

⇒

⇒ ∠PRQ = 75°

So, we have the angles as –

∠RPQ = , ∠PQR = , ∠PRQ = 75°, and ∠PRS = 105°

Option (A) is not correct because

But, ∠PRQ is not the exterior angle. So, each of the equal angles is not .

Option (C) is not correct because . But, this is not equal to the exterior angle, ∠PRS, which is 105°. So, each of the equal angles is not equal to .

Option (D) is not correct because. The value of ∠PRQ is 75° which is not the interior angle. Thus, not equal to the exterior angle. So, each of the equal angles is not equal to 75°.

Let us draw a ΔABC.

It is given to us that the angles of the triangle are in the ration 5: 3: 7.

⇒ ∠A:∠B:∠C = 5:3:7

Let us say, ∠A = 5x, ∠B = 3x, and ∠C = 7x - - - - (i)

We know that the sum of the angles of a triangle is equal to 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 5x + 3x + 7x = 180°

⇒ 15x = 180°

⇒ x = (180/15)°

⇒ x = 12° - - - - (ii)

Substituting the value of x = 12° from equation (ii) in equation (i), we get

∠A = 5x = 5 × 12° = 60°,

∠B = 3x = 3 × 12° = 36°,

∠C = 7x = 7 × 12° = 84°

So, we have ∠A = 60°, ∠B = 36°, ∠C = 84°

Since all the three angles of the triangle are less than 90°, it is an acute angled triangle.

So, option (A) is correct.

Option (B) is not correct because an obtuse angled triangle is a triangle with one obtuse angle, i.e., one angle is greater than 90° and other two angles are acute angles, i.e., the two angles are less than 90°. But, here all the three angles are less than 90°. So, it is not an obtuse angled triangle.

Option (C) is not correct, because a right triangle is a triangle with one angle equal to 90°. But, here none of the angles is equal to 90°. Instead, all the angles are less than 90°. So, it is not a right triangle.

Option (D) is not correct, because an isosceles triangle has atleast two equal sides, i.e., two equal angles. But, here all the three angles are different from each other, with no two angles equal to each other. So, it is not an isosceles triangle.

The figure for the given question is –

We have a ΔABC.

Let one of the angles of the triangle,

Let CP and BQ be the bisectors of ∠C and ∠B respectively.

Let us denote the intersection of the two bisectors, CP and BQ as D.

We have to find ∠BDC.

We know the sum of all the angles of a triangle is 180°.

So, in ΔABC,

Let us divide both the sides by 2. Then, the equation becomes

⇒ - - - - (i)

Now, in ΔBDC,

⇒

⇒ - - - - (ii)

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

⇒

⇒

⇒

⇒

⇒ (Since ∠A = 130°)

⇒

⇒

Hence, the angle between the bisectors of the two angles, i.e., ∠CDB is 155°.

Option (A) is not correct because

- - - - (iii)

Since the sum of all the angles of a triangle is 180°.

⇒

⇒ (Since ∠A = 130°)

⇒ - - - - (iv)

From equation (iii) and equation (iv),

We find that the angle between the bisectors of the two angles is not .

So, the angle between the bisectors of the two angles is not 50°.

Option (B) is not correct because

⇒

This is not the angle between the bisectors of the two angles.

Option (C) is not correct because this is not the correct angle between the bisectors of the two angles.

It is given to us that POQ is a line.

From the figure, we can say that OA and OB are two lines intersecting on the line POQ.

We know,

If a ray stands on a line, then the sum of the two adjacent angles so formed is 180°. - - - - (i)

So, ∠POA + ∠AOB + ∠BOQ = 180°

⇒ 40° + 4x + 3x = 180° (Given - ∠POA = 40°, ∠AOB = 4x, ∠BOQ = 3x)

⇒ 7x = 140°

⇒ x = 20°

Thus, option (A) is the right answer.

Option (B) is not correct because on putting x = 25° on ∠AOB = 4x and on ∠BOQ = 3x, the theorem (i) is not satisfied, i.e.,

∠POA + ∠AOB + ∠BOQ ≠180°

Thus, the value of x is not 25°.

Option (C) is not correct because on putting x = 30° on ∠AOB = 4x and on ∠BOQ = 3x, the theorem (i) is not satisfied, i.e.,

∠POA + ∠AOB + ∠BOQ ≠180°

Thus, the value of x is not 30°.

Option (D) is not correct because on putting x = 25° on ∠AOB = 4x and on ∠BOQ = 3x, the theorem (i) is not satisfied, i.e.,

∠POA + ∠AOB + ∠BOQ ≠180°

Thus, the value of x is not 25°.

In the figure, it is given to us that –

OP || RS

∠OPQ = 110°

∠QRS = 130°

We have to find ∠PQR.

Let us extend OP, so as to intersect QR at point D. The figure is as -

Now, we have OP || RS and, DR is a transversal.

We know, if a transversal intersects two parallel lines, then each pair of alternate angles are equal.

⇒ ∠DRS = ∠RDP

It is given to us that ∠QRS = 130°

⇒ ∠DRS = 130°

⇒ ∠RDP = 130° - - - - (i)

Now, we have QR as a line segment. By linear pair axiom,

∠PDQ + ∠PDR = 180°

⇒ ∠QDP + ∠RDP = 180°

⇒ ∠QDP = 180° - ∠RDP

⇒ ∠QDP = 180° - 130°

⇒ ∠QDP = 50°

⇒ ∠PDQ = 50° - - - - (ii)

In ΔPQD,

∠OPQ is an exterior angle.

Also, we know that exterior angle is equal to the sum of the two interior opposite angles.

⇒ ∠OPQ = ∠PQD + ∠PDQ - - - - (iii)

We have ∠OPQ = 110° and from equation (ii) we have ∠PDQ = 50°.

So, equation (iii) can be written as –

110° = ∠PQD + 50°

⇒ ∠PQD = 60°

So, ∠PQR = 60°

Thus, option (C) is correct.

Option (A) is not correct. If ∠PQR is equal to 40°, then it won’t satisfy the linear axiom. So, ∠PQR is not 40°.

Option (B) is not correct. We have got ∠PDQ = 50°. If ∠PQR = 50°, then the exterior angle, i.e., ∠OPQ = 110° is not equal to the sum of the opposite interior angles, ∠PDQ and ∠PQR which are now 50° each. So, ∠PQR is not equal to 50°.

Option (D) is not correct. We have, ∠OPQ = 110°. By linear pair axiom, ∠QPD = 70°. So, ∠PQR is not equal to 70⁰

Let us draw a ΔABC.

It is given that the angles of the triangle are in the ratio 2: 4: 3.

Let us assume,

∠A = 2x, ∠B = 4x, ∠C = 3x - - - - (i)

We know that the sum of the angles of a triangle is equal to 180°.

So, ∠A + ∠B + ∠C = 180°

⇒ 2x + 4x + 3x = 180° [From equation (i)]

⇒ 9x = 180°

⇒ x = 20° - - - - (ii)

From equation (ii), we get

∠A = 2x = 2 × 20° = 40°

∠B = 4x = 4 × 20° = 80°

∠C = 3x = 3 × 20° = 60°

From above, we find that the smallest angle, ∠A is 40°.

Thus, option (B) is correct.

Option (A) is not correct. The smallest angle is not 60° this is because the smallest angle is ∠A which is 40°. So, the smallest angle of the triangle is not 60°.

Option (C) is not correct. ∠B measures 80° which is the largest angle of the triangle. So, the smallest angle of the triangle is not 80°.

Option (D) is not correct. 20° is not the correct angle for any of the angles of the triangle. So, the smallest angle of the triangle is not 20°.

From the figure we can say that,

BD is a ray that intersects AB and BC at the point B which results in

∠ABD = y

and, ∠DBC = x

We know,

If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.

⇒ If the sum of two adjacent angles is 180°, then a ray stands on a line.

Thus, for ABC to be a line,

The sum of ∠ABD and ∠DBC should be equal to 180°.

⇒ ∠ABD + ∠DBC = 180°

⇒ x + y = 180°

Therefore, the value of x + y should be equal to 180° for ABC to be a line.

We know, the sum of three angles of a triangle is equal to 180°.

According to the question, if all the angles of a triangle are less than 60°, then each of the angle would be at least 59°.

⇒ Sum of the angles = 59° + 59° + 59°

⇒ Sum of the angles = 177°

⇒ The sum of the three angles of the triangle is not equal to 180°.

Similarly, for any value of the angles less than 59°, the sum of the three angles won’t be equal to 180°.

Thus, it won’t be a triangle.

We know, the sum of three angles of a triangle is equal to 180°.

Also, an obtuse angle is one whose value is greater than 90° but less than 180°.

According to the question, if the triangle has two obtuse angles, then there are at least two angles which are 91° each.

On adding these two angles,

Sum of the two angles = 91° + 91°

⇒ Sum of the two angles = 182°

This already exceeds the sum of three angles of the triangle, even without considering the third angle.

Again, if we consider the case where one angle of the triangle is 90°, we have a right - angled triangle. The rest two angles must be less than 90° each so as to satisfy the property of a triangle, which states that the sum of the three angles of a triangle should be equal to 180°.

Thus, a triangle cannot have two obtuse angles.

It is given that the triangle has its angles as 45°, 64°, and 72°.

On adding these, we get

Sum of three angles = 45° + 64° + 72°

⇒ Sum of three angles = 181°

According to linear pair of axiom, the sum of three angles of the triangle should be equal to 180°.

Since, the sum of the given angles exceeds 180°, no such triangle exists with its angles as 45°, 64°, and 72°.

It is given that the triangle has its angles as 53°, 64°, and 63°.

Upon adding these, we get

⇒ Sum of the three angles = 53° + 64° + 63°

⇒ Sum of the three angles = 180° - - - - (i)

We know, that the sum of all the angles of a triangle is equal to 180°.

Also, from (i), we find out that the sum of the given angles of the triangle is equal to 180°.

Thus, infinite number of triangles can be drawn having angles as 53°, 64°, and 63°.

Let us draw the figure as below -

It is given to us that l and m are parallel to each other.

Here, n is a transversal intersecting l and m which are parallel to each other.

Also, we have ∠pqm = 44° - - - - (i)

We have to find the value of x, i.e., ∠qpl

We know, if a transversal intersects two parallel lines then each pair of corresponding angles is equal.

Here, the transversal n intersects two parallel lines l and m. So, the following holds true for the corresponding angles.

∠pqm = ∠npl

⇒ ∠npl = 44° (From (i), we have ∠pqm = 44°) - - - - (ii)

Again, the linear pair axiom states that

If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.

Here, we can see that l is a ray standing on the line n.

⇒ ∠npl + ∠lpq = 180° (By linear pair axiom)

⇒ 44° + ∠lpq = 180°

⇒ ∠lpq = 180° - 44°

⇒ ∠lpq = 136°

⇒ x = 136°

Thus, the value of x is equal to 136°.

We know,

Two angles are said to be adjacent, if they have a common vertex, a common arm, and their non - common arms are situated on different sides of the common arm.

Also, we know that the linear pair axiom states that, if a ray stands on a line, then the sum of the two adjacent angles so formed is 180°.

It is given to us that the two adjacent angles are equal. Let us assume each of the adjacent angles measures x.

We know that the sum of two adjacent angles is 180°.

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = 90°.

Thus, each of the adjacent angle is 90°.

Let us draw two intersecting lines, AB and CD.

Let us denote the intersection point of AB and CD as point P.

It is given to us that one of the angles formed by these two intersecting lines, is a right angle, i.e., 90°.

Let us assume ∠CPA is equal to 90°. - - - - (i)

Now, CPD is a straight line and by linear pair axiom,

∠CPA + ∠APD = 180°

⇒ 90° + ∠APD = 180° [From equation (i)]

⇒ ∠APD = 180° - 90°

⇒ ∠APD = 90° - - - - (ii)

Similarly, APB is a straight line and by linear pair axiom,

∠APD + ∠DPB = 180°

⇒ 90° + ∠DPB = 180°

⇒ ∠DPB = 180° - 90°

⇒ ∠DPB = 90° - - - - (iii)

In the same way, ∠CPA + ∠BPC = 180°

⇒ 90° + ∠BPC = 180°

⇒ ∠BPC = 180° - 90°

⇒ ∠BPC = 90° - - - - (iv)

From equations (ii), (iii), and (iv), we find that the rest of the three angles are also 90° each.

Thus, all the four angles are 90° each.

In the 1^st figure –

Let the intersecting points of l and n be a, and of m and n be b respectively.

We have, ∠lab = 132°, ∠abm = 48° - - - - (i)

l is a ray on the line n. By linear pair axiom,

∠nal + ∠lab = 180°

⇒ ∠nal + 132° = 180°

⇒ ∠nal = 180° - 132°

⇒ ∠nal = 48° - - - - (ii)

Also, we have ∠abm = 48° from equation (i)

We can say that, the two corresponding angles,

∠abm = ∠nal = 48° - - - - (iii)

Since, the two corresponding angles are equal, such that a transversal n intersects two lines, then they are parallel to each other.

The following figure can be drawn from the given details.

Here, we have

l and m are two lines perpendicular to the same line n.

⇒ ∠lan = ∠mbn = 90° - - - - (i)

Now, we can see that n is a transversal intersecting two lines l and m at points a and b, such that,

the corresponding angles, ∠lan = ∠mbn and are 90° each.

Since, the corresponding angles between two lines are equal when a transversal intersects these two lines at two different points, the two lines are parallel to each other.

Thus, the lines l and m are parallel to each other, not perpendicular.

From the figure, it is given to us

OD⊥OE

⇒ ∠DOE = 90° - - - - (i)

OD is the bisector of ∠AOC.

OE is the bisector of ∠BOC.

We have to show that the points A, O and B are collinear, i.e.,

To show that AOB is a straight line. - - - - (ii)

OD is the bisector of ∠AOC

⇒ ∠AOD = ∠COD

⇒ ∠AOC = 2 × ∠COD - - - - (ii)
Similarly, OE is the bisector of ∠BOC

⇒ ∠BOE = ∠COE

⇒ ∠BOC = 2 × ∠COE - - - - (iii)

Adding equation (ii) and equation (iii),

∠AOC + ∠BOC = 2 × ∠COD + 2 × ∠COE

⇒ ∠AOC + ∠BOC = 2 × (∠COD + ∠COE)

From the figure, we can see that

∠COD + ∠COE = ∠DOE - - - - (iv)

Substituting equation (iv) in the above equation,

∠AOC + ∠BOC = 2 × (∠COD + ∠COE)

⇒ ∠AOC + ∠BOC = 2 × ∠DOE

⇒ ∠AOC + ∠BOC = 2 × 90° (From equation (i), we have ∠DOE = 90°)

⇒ ∠AOC + ∠BOC = 180°

⇒ ∠AOB = 180° (From the figure, ∠AOC + ∠BOC = ∠AOB) - - - - (iv)

From (iv), we can say that

∠AOC and ∠BOC are forming linear pair of angles.

Since, ∠AOC and ∠BOC are two adjacent angles and their sum is 180°, where OC is a ray standing on AOB,

It is true that A, O and B are collinear, thus making AOB a straight line.

It is given to us –

∠1 = 60°

∠6 = 120°

We have to show that m and n are parallel to each other.

We can see that l is a ray standing on the line m. So, by linear pair axiom,

∠1 + ∠4 = 180°

⇒ 60° + ∠4 = 180°

⇒ ∠4 = 180° - 60°

⇒ ∠4 = 120° - - - - (i)

Similarly,

∠1 + ∠2 = 180°

⇒ 60° + ∠2 = 180°

⇒ ∠2 = 180° - 60°

⇒ ∠2 = 120° - - - - (ii)

Again, ∠2 + ∠3 = 180°

⇒ 120° + ∠3 = 180°

⇒ ∠3 = 180° - 120°

⇒ ∠3 = 60° - - - - (iii)

Since, ∠6 = 120° and ∠2 = 120° [from equation (ii)],

We can say that these corresponding angles are equal, i.e.,

∠6 = ∠2 = 120° - - - - (iv)

We can say that l is a ray standing on the line n. By linear pair axiom,

∠6 + ∠5 = 180°

⇒ 120° + ∠5 = 180°

⇒ ∠5 = 180° - 120°

⇒ ∠5 = 60° - - - - (v)

Since, ∠1 = 60° and ∠5 = 60° [from equation (v)],

We can say that these corresponding angles are equal, i.e.,

∠1 = ∠5 = 60° - - - - (vi)

Similarly, we get

∠8 = ∠4 = 120° (which are also the corresponding angles, and from equation (i), ∠4 = 120°)

And, ∠7 = ∠3 = 60° (which are also the corresponding angles, and from equation (iii), we have ∠3 = 60°)

Thus, we can say that

l is a transversal intersecting two lines m and n such that each pair of corresponding angles are equal.

Then, lines m and n are parallel to each other.

It is given to us –

l and m are two parallel lines.

t is a transversal intersecting l and m at A and B respectively.

Here, ∠EAB and ∠ABH, ∠FAB and ∠GBA, are a pair of alternate interior angles which are equal to each other.

⇒ ∠EAB = ∠ABH, and ∠FAB = ∠GBA - - - - (i)

AP and BQ are the bisectors of the angles ∠EAB and ∠ABH respectively.

⇒ ∠EAP = ∠PAB, and ∠ABQ = ∠QBH - - - - (ii)

From equation (i), we have

∠EAB = ∠ABH

⇒ ∠EAP + ∠PAB = ∠ABQ + ∠QBH (From the figure)

⇒ 2 × ∠PAB = 2 × ∠ABQ [From equation (ii)]

⇒ ∠PAB = ∠ABQ - - - - (iii)

Now, t is a transversal intersecting two lines AP and BQ at A and B respectively.

⇒ ∠PAB and ∠ABQ are a pair of alternate interior angles.

Thus, from equation (iii) we can say that ∠PAB and ∠ABQ are a pair of alternate interior angles, which are equal to each other.

Thus, it is true that AP || BQ.

It is given to us –

∠EAB and ∠ABH are a pair of alternate interior angles.

AP and BQ are the bisectors of the angles ∠EAB and ∠ABH respectively.

⇒ ∠EAP = ∠PAB, and ∠ABQ = ∠QBH - - - - (i)

Also, AP || BQ

We have to show that l || m.

Now, since t is a transversal intersecting two parallel lines AP and BQ at points A and B,

∠PAB = ∠ABQ (alternate interior angles) - - - - (ii)

Now,

∠EAB = ∠EAP + ∠PAB, and ∠ABH = ∠ABQ + ∠QBH

⇒ ∠EAB = 2 × ∠PAB, and ∠ABH = 2 × ∠ABQ [From equation (i)]

⇒ ∠EAB = 2 × ∠PAB, and ∠ABH = 2 × ∠PAB

⇒ ∠EAB = ∠ABH - - - - (iii)

From the figure and the equation (iii), we can say that –

t is a transversal intersecting lines l and m such that the alternate angles, ∠EAB and ∠ABH are equal.

Thus, it is true that l || m.

It is given to us –

BA || ED

BC || EF

We have to show that ∠ABC = ∠DEF

Let us extend DE to intersect BC at P. Then, the figure becomes –

Since, BA || ED

⇒ BA || DP

We have two parallel lines BA and DP, and BP is a transversal intersecting BA and DP at points B and P respectively.

⇒ ∠ABC = ∠DPC (corresponding angles) - - - - (i)

Similarly, we have two parallel lines BC and EF, and DP is a transversal intersecting BC and EF at points E and P respectively.

⇒ ∠DPC = ∠DEF (corresponding angles) - - - - (ii)

From (i) and (ii), we have

∠ABC = ∠DPC, and ∠DPC = ∠DEF

⇒ ∠ABC = ∠DEF

Hence, proved.

It is given to us –

BA || ED
BC || EF

To show that - ∠ABC + ∠DEF = 180°

Let us extend DE to intersect BC at G, and EF to intersect BA at H. Then, the figure becomes –

Since, BA || DE

⇒ BA || GE

We have two parallel lines BA and GE, and BG is a transversal intersecting BA and GE at points B and G respectively.

⇒ ∠ABC = ∠EGC - - - - (i)

Also, BC || EF, and GE is a transversal intersecting BC and EF at points G and E respectively.

⇒ ∠EGC = ∠HEG - - - - (ii)

Since GE is a ray standing on the line HF. By linear pair axiom,

∠HEG + ∠GEF = 180°

⇒ ∠EGC + ∠GEF = 180° [From equation (ii)]

⇒ ∠ABC + ∠GEF = 180°

⇒ ∠ABC + ∠DEF = 180°

Hence, proved.

It is given to us –

DE || QR and n is the transversal intersecting DE and QR at points A and B respectively.

We have to find ∠APB.

AP and BP are the bisectors of ∠EAB and ∠ABR respectively.

⇒ ∠EAP = ∠PAB

⇒ ∠EAB = 2 × ∠PAB - - - - (i)

Also, ∠RBP = ∠PBA

⇒ ∠RBA = 2 × ∠PBA - - - - (ii)

Since, DE || QR,

∠RBA = ∠EAn (Corresponding angles)

⇒ ∠RBA = 180° - ∠EAB (By linear pair axiom, ∠EAn + ∠EAB = 180°)

Using equations (i) and (ii) in the above equation,

2 × ∠PBA = 180° - (2 × ∠PAB)

⇒ ∠PBA = 90° - ∠PAB (Dividing both sides by 2)

⇒ ∠PBA + ∠PAB = 90° - - - - (iii)

Now, in ΔPAB,

∠PBA + ∠PAB + ∠APB = 180° (Sum of three angles of a triangle is 180°)

⇒ 90° + ∠APB = 180° [From equation (iii)]

⇒ ∠APB = 90°

Thus, the value of ∠APB is 90°.

It is given to us –

The angles of a triangle are in the ratio 2: 3: 4.

We have to find the angles of the triangle.

Let us draw a ΔABC. Since the angles of the triangle is in the ratio 2: 3: 4, we can say that

∠A:∠B:∠C = 2:3:4

Let us assume,

∠A = 2x, ∠B = 3x, and ∠C = 4x - - - - (i)

We know that the sum of the three angles of a triangle is equal to 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 2x + 3x + 4x = 180° [From equation (i)]

⇒ 9x = 180°

⇒ x = 20°

Substituting the value of x = 20° in equation (i), we get

∠A = 2x = 2 × 20° = 40°,

∠B = 3x = 3 × 20° = 60°, and

∠C = 4x = 4 × 20° = 80°

Thus, the angles of the triangle are 40°, 60°, and 80°.

It is given to us –

ΔABC is a right - angled triangle.

∠A = 90°

L is a point on BC

AL ⊥ BC

We have to prove ∠BAL = ∠ACB.

We know that the sum of the angles of a triangle is equal to 180°. Thus, in ΔABC,

∠BAC + ∠B + ∠C = 180° - - - - (i)

Now, in ΔABL,

AL ⊥ BC, i.e., ∠ALB = 90°

Since, the sum of the angles of a triangle is equal to 180°,

∠BAL + ∠ALB + ∠B = 180° - - - - (ii)

From equation (i) and equation (ii), we can say that

∠BAC + ∠B + ∠C = ∠BAL + ∠ALB + ∠B

⇒ ∠C + ∠BAC = ∠BAL + ∠ALB

⇒ ∠C = ∠BAL (Since, ∠A = 90° and ∠ALB = 90°, so they are equal)

⇒ ∠BAL = ∠ACB

Hence, proved.

Let us draw the figure as below –

It is given to us that there are two parallel lines. Let us assume l and m are two parallel lines, i.e., l || m.

Also, it is given that two lines are perpendicular to the parallel lines. Let us assume n and p are the two perpendicular lines.

⇒ n ⊥ l, n ⊥ m, p ⊥ l, and p ⊥ m

Let us assume the angles as ∠1, ∠2, …, ∠16 as shown in the figure.

To show that n and p are parallel to each other, i.e., to prove n || p.

We know that l || m, n ⊥ l, and n ⊥ m.

⇒ ∠1 = ∠2 = ∠3 = ∠4 = 90°, and ∠9 = ∠10 = ∠11 = ∠12 = 90°

⇒ ∠1 = ∠9 = 90°, ∠2 = ∠10 = 90° (Corresponding angles)

Similarly, ∠4 = ∠10 = 90°, ∠3 = ∠9 = 90° (Alternate interior angles)

Again, l || m, p ⊥ l, and p ⊥ m.

⇒ ∠5 = ∠6 = ∠7 = ∠8 = 90°, and ∠13 = ∠14 = ∠15 = ∠16 = 90°

We have ∠3 = 90° and ∠8 = 90°

⇒ ∠3 + ∠8 = 90° + 90° = 180°

Thus, we can say that the sum of the two interior angles is supplementary.

We know, if a transversal intersects two lines, such that each pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel to each other.

Thus, n || p.

Let us draw the figure.

Here, we can see that

AB and CD intersect each other at point E.

The two pairs of vertically opposite angles are -

1^st pair - ∠AEC and ∠BED

2^nd pair - ∠AED and ∠BEC

We need to prove that the vertically opposite angles are equal, i.e.,

∠AEC = ∠BED, and ∠AED = ∠BEC

Now, we can see that the ray AE stands on the line CD. We know, if a ray stands on a line then the sum of the adjacent angles is equal to 180°.

⇒ ∠AEC + ∠AED = 180° (By linear pair axiom) - - - - (i)

Similarly, the ray DE stands on the line AEB.

⇒ ∠AED + ∠BED = 180° (By linear pair axiom) - - - - (ii)

From equations (i) and (ii), we have

∠AEC + ∠AED = ∠AED + ∠BED

⇒ ∠AEC = ∠BED - - - - (iii)

Similarly, the ray BE stands on the line CED.

⇒ ∠DEB + ∠CEB = 180° (By linear pair axiom) - - - - (iv)

Also, the ray CE stands on the line AEB.

⇒ ∠CEB + ∠AEC = 180° (By linear pair axiom) - - - - (v)

From equations (iv) and (v), we have

∠DEB + ∠CEB = ∠CEB + ∠AEC

⇒ ∠DEB = ∠AEC - - - - (vi)

Thus, from equation (iii) and equation (vi), we have

∠AEC = ∠BED, and ∠DEB = ∠AEC

Therefore, it is proved that the vertically opposite angles are equal.

Let us draw the figure as below –

Here, we have

A ΔABC

BC is extended to D.

Let BT be the bisector of ∠B of the triangle.

Also, let us assume the bisector of ∠ACD to be CT.

It is given that BO and CT intersect at point T.

We have to prove that ∠BTC = 1/2∠BAC

We know, if a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles. Here, ∠ACD is an exterior angle and the two interior opposite angles are ∠ABC and ∠CAB.

⇒ ∠ACD = ∠ABC + ∠CAB

Dividing both sides of the equation by 2,

1/2∠ACD = 1/2∠ABC + 1/2∠CAB

⇒ ∠TCD = 1/2∠ABC + 1/2∠CAB (∠TCD = 1/2∠ACD, since CT is the bisector of ∠ACD) - - - - (i)

Again, in ΔBTC, ∠TCD is an exterior angle of the triangle and the two opposite interior angles are ∠BTC and ∠CBT.

⇒∠TCD = ∠BTC + ∠CBT

⇒ ∠TCD = ∠BTC + 1/2∠ABC (∠CBT = 1/2∠ABC, since BT is the bisector of ∠ ABC) - - - - (ii)

From equations (i) and (ii), we can say that

1/2∠ABC + 1/2∠CAB = ∠BTC + 1/2∠ABC

⇒ 1/2∠CAB = ∠BTC

⇒ ∠BTC = 1/2 ∠CAB

⇒ ∠BTC = 1/2 ∠BAC

Hence, proved.

Let us draw the figure as below –

The two parallel lines are PR and QT.

Line A intersects P and Q at points C and D respectively.

Let CB and DE be the bisectors of ∠ACR and ∠ADT respectively.

We have to prove that CB and DE are parallel to each other.

We know, if two lines are parallel to each other, the corresponding angles are equal.

⇒

Dividing both sides by 2,

⇒

⇒

Now, we have two lines CB and DE such that the corresponding angles, ∠ACB and ∠CDE are equal.

Thus, CB || DE

Therefore, it is proved that the bisectors of any pair of corresponding angles so formed are parallel.

Let us draw a line m. Let A be a point.

Let us draw two intersecting lines l and p through the point A, such that they are perpendicular to the line m at points P and Q.

⇒ ∠P = 90°, and ∠Q = 90° - - - - (i)

We have to prove that through a given point, we can draw only one perpendicular to a given line.

⇒ To prove that ∠A = 0°

Now, we can see in ΔPAQ,

∠P + ∠A + ∠Q = 180° (Since the sum of the angles of a triangle is equal to 180°)

⇒ ∠A + 90° + 90° = 180° [From equation (i)]

⇒ ∠A + 180° = 180°

⇒ ∠A = 0°

This means that ∠A doesn’t exist, i.e., the lines l and p coincide with each other.

Therefore, it is true that through a given point, we can draw only one perpendicular to a given line.

Let us draw the figure as below –

It is given to us that

Two lines are intersecting each other. Let us assume l and m to be the two intersecting lines.

Also, we have two lines that are perpendicular to the two intersecting lines. Let us say, a ⊥ l, and b ⊥ m.

We have to prove that a and b intersect each other.

Let us assume that a and b do not intersect.

⇒ a || b

Now, we have a ⊥ l and, a || b

⇒ b ⊥ l - - - - (i)

Also, we have b ⊥ m - - - - (ii)

From (i) and (ii), we can say that this situation will hold true if and only if l || m.

But, this is incorrect because it is given to us that l and m are two intersecting lines.

Hence, our initial assumption is wrong.

Thus, a and b intersect each other.

Therefore, it is proved that two lines that are respectively perpendicular to two intersecting lines intersect each other.

Let us draw a ΔABC as below:

We have to prove that a triangle must have at least two acute angles.

Let us assume a case where two angles are right angles, i.e., 90° each.

Let these angles be ∠B and ∠C

⇒ ∠B = 90°, and ∠C = 90°

We know that the sum of the three angles of a triangle is equal to 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + 90° = 180° (Since, ∠B = 90°, and ∠C = 90°)

⇒ ∠A = 0°, which is not possible because then, no triangle would exist.

Thus, two angles of a triangle cannot be 90° each.

Let us assume another case where two angles of the triangle are obtuse angle, i.e., each of the angles is more than 90°.

Let the obtuse angles be ∠B and ∠C

⇒ ∠B + ∠C>180° because each of them is more than 90°.

We know that the sum of the three angles of a triangle is equal to 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A = 180° - (∠B + ∠C)

⇒ ∠A = a negative value, which is not possible.

Thus, no such triangle is possible which has its two angles greater than 90°.

Again, let us assume a case where one angle is 90° and another angle is an obtuse angle, i.e., greater than 90°.

Let’s say ∠B = 90°, and ∠C is obtuse, i.e., ∠C > 90°.

We know that the sum of the three angles of a triangle is equal to 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A = 180° - 90° - ∠C

⇒ ∠A = 90° - ∠C

Since, ∠C > 90°, the value of ∠A becomes negative.

Thus, no such triangle exists such that one angle is 90° and other angle is an obtuse angle.

Let us assume the case when two angles are acute, i.e., both the angles are less than 90°.

Let these angles be ∠B and ∠C.

⇒ The sum of these two angles is less than 180°.

⇒ ∠B + ∠C<180°

We know that the sum of the three angles of a triangle is equal to 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A = 180° - (∠B + ∠C)

⇒ ∠A = a positive value, since ∠B + ∠C<180°

⇒ ∠A = an acute angle

Thus, it is proved that a triangle should have atleast two acute angles.

It is given to us –

∠Q > ∠R

PA is the bisector of ∠QPR

⇒ ∠QPA = ∠RPA - - - - (i)

PM ⊥ QR

⇒ ∠PMR = ∠PMQ = 90° - - - - (ii)

We have to prove that ∠APM = 1/2 (∠Q - ∠R)

Since, the sum of the three angles of a triangle is equal to 180°, in ΔPQM,

∠PQM + ∠PMQ + ∠QPM = 180°

⇒ ∠PQM + 90° + ∠QPM = 180° [From equation (ii)]

⇒ ∠PQM + ∠QPM = 180° - 90°

⇒ ∠PQM + ∠QPM = 90°

⇒ ∠PQM = 90° - ∠QPM - - - - (iii)

Similarly, in ΔPMR, the sum of the three angles of a triangle is equal to 180°

⇒ ∠PMR + ∠PRM + ∠RPM = 180°

⇒ 90° + ∠PRM + ∠RPM = 180° [From equation (ii)]

⇒ ∠PRM + ∠RPM = 180° - 90°

⇒ ∠PRM + ∠RPM = 90°

⇒ ∠PRM = 90° - ∠RPM - - - - (iv)

Now, on subtracting equation (iv) from equation (iii), we get

∠PQM - ∠PRM = (90° - ∠QPM) - (90° - ∠RPM) ×

⇒ ∠Q - ∠R = 90° - ∠QPM - 90° + ∠RPM

⇒ ∠Q - ∠R = ∠RPM - ∠QPM

⇒ ∠Q - ∠R = (∠RPA + ∠APM) - (∠QPA - ∠APM)

⇒ ∠Q - ∠R = ∠RPA + ∠APM - ∠QPA + ∠APM

⇒ ∠Q - ∠R = ∠QPA + 2 × ∠APM - ∠QPA [From equation (i)]

⇒ ∠Q - ∠R = 2 × ∠APM

⇒ ∠APM = 1/2 (∠Q - ∠R)

Hence, proved.