Triangles

It is given in the question that:

AC = AD and,

AB bisects ∠A

To prove:

Proof: In

AB = AB (Common)

AC = AD (Given)

∠CAB = ∠DAB (AB is bisector)

Therefore,

By SAS congruence,

BC and BD are of equal length by C.P.C.T.

It is given in the question that:

AD = BC and

∠DAB = ∠CBA

(i) In ΔABD and ΔBAC,

AB = BA (Common)

∠DAB = ∠CBA (Given)

AD = BC (Given)

Therefore By SAS congruence

ΔABD ≅ ΔBAC

(ii) Since,

ΔABD ≅ ΔBAC

Therefore,

BD = AC (By c.p.c.t)

(iii) Since,

ΔABD ≅ ΔBAC

Therefore,

∠ABD = ∠BAC (By c.p.c.t)

It is given in the question that:

AD and BC are equal perpendiculars to AB

To prove: CD bisects AB

Proof: In ΔAOD and ΔBOC

∠A = ∠B (Perpendicular)

∠AOD = ∠BOC (Vertically opposite angle)

AD = BC (Given)

Therefore,

By AAS congruence,

ΔAOD ≅ ΔBOC

Now,

AO = OB (By c.p.c.t)

CD bisects AB

It is given in the figure given,

l parallel to m

And, p parallel to q

To prove:- ΔABC ≅ ΔCDA
In ΔABC and ΔCDA

∠BCA = ∠DAC (Alternate interior angles)

∠BAC = ∠DCA (Alternate interior angles)

AC = AC (common)

Two angles and one side in between are equal, therefore, By ASA congruence,

ΔABC ≅ ΔCDA

Given: l is the bisector of an ∠A, BP and BQ are perpendiculars

(i)
To Prove:
Proof:
In

∠P = ∠Q (Right angles)

∠BAP = ∠BAQ (l is the bisector)

AB = AB (Common)

Therefore,

By AAS congruence,

(ii) BP = BQ (By c.p.c.t)

Therefore,

B is equidistant from the arms of ∠A

Given:

AC = AE

AB = AD and,

∠BAD = ∠EAC

To show: BC = DE

Proof:

∠BAD = ∠EAC
Adding ∠DAC both sides

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠EAD

In

AC = AE (Given)

∠BAC = ∠EAD

AB = AD (Given)

Therefore,

By SAS congruence,

BC = DE (By c.p.c.t)

It is given in the question that:

P is the mid-point of AB

∠BAD = ∠ABE and,

∠EPA = ∠DPB

(i) ∠EPA = ∠DPB
Adding ∠DPE both sides

∠EPA + ∠DPE = ∠DPB + ∠DPE

∠DPA = ∠EPB

In

∠DPA = ∠EPB

AP = BP (P is the mid-point of AB)

∠BAD = ∠ABE (Given)

Therefore,

By ASA congruence,

(ii) AD = BE (By c.p.c.t)

Given: ∠ACB = 90^o, M is the mid-point of AB and DM = CM

(i) In

AM = BM (M is the mid-point)

∠CMA = ∠DMB (Vertically opposite angle)

CM = DM (Given)

Therefore by SAS congruency.

(ii) ∠ACM = ∠BDM (By c.p.c.t)

Therefore,

AC parallel to BD as alternate interior angles are equal

Now,

∠ACB + ∠DBC = 180^o (Co-interior angles)

90^o + ∠B = 180^o

∠DBC = 90^o

(iii) In

BC = CB (Common)

∠ACB = ∠DBC (Right angles)

DB = AC (By c.p.c.t already proved)

By SAS congruence rule,

(iv) DC = AB ()

DM = CM = AM = BM (M is the mid-point)

DM + CM = AM + BM

CM + CM = AB

CM = AB

Given: AB = AC

The bisectors of ∠B and ∠C intersect each other at O

(i)
To Prove = OB = OC
Proof:
ABC is an isosceles with AB = AC

Therefore,

∠B = ∠C

Multiplying both sides by 1/2, we get,

∠B = ∠C (Bisectors of angles are also be equal)

∠OBC = ∠OCB (Angles bisectors)

OB = OC (Side opposite to the equal angles are equal)

(ii)

To Prove: AO bisects angle A

Proof:

In

AB = AC (Given)

AO = AO (Common)

OB = OC (Proved above)

Therefore,

By SSS congruence rule

∠BAO = ∠CAO (By c.p.c.t)

Thus,

AO bisects ∠A

Given: AD is the perpendicular bisector of BC

To show: AB = AC

Proof: In

AD = AD (Common)

∠ADB = ∠ADC

BD = CD (AD is the perpendicular bisector)

Therefore,

By SAS congruence axiom,

[CPCT: "Corresponding parts of congruent triangles are congruent"]

AB = AC (By c.p.c.t)

Given: BE and CF are altitudes, AC = AB

To show: BE = CF

Proof:
In

∠A = ∠A (Common)

∠AEB = ∠AFC (Right angles)

AB = AC (Given)

AAS Postulate (Angle-Angle-Side) If two angles and a non-included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.
Therefore,
By AAS congruence axiom,

Thus,

BE = CF (By corresponding parts of congruent triangles)
Hence, Proved.

It is given in the question that:

BE = CF

(i)
To Prove:
Proof:
In

∠A = ∠A (Common)

∠AEB = ∠AFC (Right angles)

BE = CF (Given)

Therefore,

By AAS congruence rule,

(ii) Thus,

AB = AC (By c.p.c.t)

Therefore,

ABC is an isosceles triangle

It is given in the question that ABC and DBC are two isosceles triangles

To show: ∠ABD = ∠ACD

Proof: Let us join AD, as shown in the figure:

In, ΔABD and ΔACD

AD is the common side

AB = AC (ABC is an isosceles triangle)

BD = CD (BCD is an isosceles triangle)

Therefore,

By Side-Side-Side Congruency,

ΔABD ≅ ΔACD

Thus,

∠ABD = ∠ACD (By Corresponding Parts of Congruent Triangles)

Given: AB = AC and AD = AB
To Prove: ∠ BCD is a right angle

Proof: In

AB = AC (Given)

∠ACB = ∠ABC (Angles opposite to equal sides are equal)

In

AD = AB (Given)

∠ADC = ∠ACD (Angles opposite to equal sides are equal)

Now,

In

∠CAB + ∠ACB + ∠ABC = 180^o ( Sum of interior angles of a triangle)

∠CAB + 2 ∠ACB = 180^o (∠ACB = ∠ABC)

∠CAB = 180^o - 2 ∠ACB (i)

Similarly,

In

∠CAD = 180^o - 2 ∠ACD (ii)

Also,

∠CAB + ∠CAD = 180^o (BD is a straight line)

Adding (i) and (ii), we get

∠CAB + ∠CAD = 180^o - 2∠ACB + 180^o - 2∠ACD

180^o = 360^o - 2∠ACB - 2∠ACD

2 (∠ACB + ∠ACD) = 180^o

∠BCD = 90^o
Hence, Proved.

It is given in the question that:

∠A = 90^o

And,

AB = AC

According to question,

AB = AC

∠B = ∠C (Angles opposite to equal sides are equal)

Now,

∠A + ∠B + ∠C = 180^o (Sum of the interior angles of a triangle)

90^o + 2∠B = 180^o

2∠B = 90^o

∠B = 45^o

Thus,

∠B = ∠C = 45^o

Let us consider,

ABC be an equilateral triangle

BC = AC = AB (Length of all sides is same)

∠A = ∠B = ∠C (Sides opposite to the equal angles are equal)

Also,

∠A + ∠B + ∠C = 180^o ( Sum of interior angles of a triangle is 180°)

3∠A = 180^o

∠A = 60^o

Therefore,

∠A = ∠B = ∠C = 60^o

Thus,

The angles of an equilateral triangle are 60^o each

It is given in the question that:

are two isosceles triangles

(i) In ,

AD = AD (Common)

AB = AC (Triangle ABC is isosceles)

BD = CD (Triangle DBC is isosceles)

Therefore,

By SSS axiom,

(ii) In

AP = AP (Common)

∠PAB = ∠PAC (By c.p.c.t)

AB = AC (Triangle ABC is isosceles)

Therefore,

By SAS axiom,

(iii) ∠PAB = ∠PAC (By c.p.c.t)

AP bisects ∠A (i)

Also,

In

PD = PD (Common)

BD = CD (Triangle DBC is isosceles)

BP = CP ( so by c.p.c.t)

Therefore,

By SSS axiom,

∠BDP = ∠CDP (By c.p.c.t) (ii)

By (i) and (ii), we can say that AP bisects ∠A as well as ∠D

(iv) ∠BPD = ∠CPD (By c.p.c.t)

And,

BP = CP (i)

Also,

∠BPD + ∠CPD = 180^o (BC is a straight line)

2∠BPD = 180^o

∠BPD = 90^o (ii)

From (i) and (ii), we get

AP is the perpendicular bisector of BC

It is given in the question that:

AD is an altitude and AB = AC

(i) In

∠ADB = ∠ADC = 90^o

AB = AC (Given)

AD = AD (Common)

Therefore,

By RHS axiom,

(By c.p.c.t)

(ii) ∠BAD = ∠CAD (By c.p.c.t)

Thus,

AD bisects ∠A

It is given in the question that:

AB = PQ

BC = QR

And,

AM = PN

(i) To Prove:
BC = BM and QR = QN (AM and PN are medians)

BC = QR

BC = QR

In

AM = PN (Given)

AB = PQ (Given)

BM = QN (Proved above)

Therefore,

By SSS congruence rule,

(ii)
To Prove:
Proof:
In and

AB = PQ (Given)

∠ABC = ∠PQR (By c.p.c.t)

BC = QR (Given)

Therefore,

By SAS congruence rule,

Given: BE and CF are two equal altitudes

In

∠BEC = ∠CFB = 90^o (Angle made by Altitudes)

BC = CB (Common)

BE = CF (Given equal altitudes)

Therefore,

By RHS axiom,

Now,

∠C = ∠B (By c.p.c.t)

Thus,

AB = AC as sides opposite to the equal angles are equal
Thus triangle ABC is an isosceles triangle.
Hence, Proved.

It is given in the question that:

AB = AC

In and

∠APB = ∠APC = 90^o (AP is altitude)

AB = AC (Given)

AP = AP (Common)

Therefore,

By RHS axiom,

Thus,

∠B = ∠C (By c.p.c.t)

ABC is a triangle right angled at A.

Now,

∠A + ∠B + ∠C = 180^o ( Sum of interior angles of a triangle)

∠B + ∠C = 90^o
So, ∠B and ∠C must be less than 90^o

And ∠A is 90^o
Since side opposite to largest angle is the largest and Since, "A" is the largest angle of the triangle, the side opposite to it must be the largest

So, BC is the hypotenuse which is the largest side of the right-angled triangle ABC.

It is given in the question that:

∠PBC < ∠QCB

Now,

∠ABC + ∠PBC = 180^o (linear pair)

∠ABC = 180^o - ∠PBC

Also,

∠ACB + ∠QCB = 180^o(linear pair)

∠ACB = 180^o - ∠QCB

Since,

∠PBC < ∠QCB therefore,

∠ABC > ∠ACB

Thus,

AC > AB as sides opposite to the larger angle is larger.

Given: ∠B < ∠A and, ∠C < ∠D

Now,

AO < BO (i) (Side opposite to the smaller angle is smaller)

OD < OC (ii) (Side opposite to the smaller angle is smaller)

Adding (i) and (ii), we get

AO + OD < BO + OC

AD < BC

It is given in the question that:
AB = shortest side

In ΔADC,

AD<CD (CD is the largest side of ABCD)

∴ ∠4 < ∠3 (Angle Opposite to smaller side is smaller) (1)

In ΔABC,

AB<BC (AB is smallest side of quadrilateral ABCD)

∴ ∠2 < ∠1 (Angle opposite to smaller side is smaller) (2)

Adding (1) and (2) gives,

∠2 + ∠4 < ∠1 + ∠3

⇒ ∠C < ∠A

⇒ ∠A > ∠C

Now, in ΔABD,

AB < AD (AB is smallest side of ABCD)

∴ ∠8 < ∠5 (Angle opposite to smaller side is smaller) (3)

In ΔBDC,

BC < CD (CD is longest side)

∴ ∠7 < ∠6 (Angles opposite to smaller side is smaller) (4)

On adding (3) and (4), we have,

∠8 + ∠7 < ∠5 + ∠6

⇒ ∠D < ∠B

⇒ ∠B > ∠D

Hence proved.

Given: PR > PQ and PS bisects ∠QPR

To prove: ∠PSR > PSQ

Proof:
Given, PR > PQ,

∴ ∠PQR > ∠PRQ (Angles opp. to the longer side is greater)

PS is the bisector of ∠QPR.

∴ ∠QPS = ∠RPS

Let ∠QPS = ∠RPS = y

In ΔPQS,

∠RPS is the exterior angle.

We know that exterior angle is sum of interior opp. angles

∴ ∠PSR = ∠PQR + y …. (1)

In ΔPSR,

∠PSQ is the exterior angle.

∴ ∠PSQ = ∠PRQ + y …. (2)

Now,

∠PQR > ∠PRQ

Adding y on both sides to get,

∠PQR + y> ∠PRQ + y

∠PSR > ∠PSQ
Hence Proved.

Let l is a line segment and B is a point lying to it. We draw a line AB perpendicular to l. Let C be any other point on l

To prove: AB < AC

Proof: In ,

∠B = 90^o

Now,

∠A + ∠B + ∠C = 180^o

∠A + ∠C = 90^o

Therefore,

∠C must be an acute angle or ∠C < ∠B

AB < AC (Sides opposite to the larger angle is larger)

Or, AB is shortest distance.

Circum centre of a triangle is always equidistant from all the vertices of that particular triangle.

Circum centre is the point where perpendicular bisectors of all the sides of the triangle meet together.

In ∆ABC, we can find the circum centre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle.

O is the point where these bisectors are

meeting together.

Therefore,

O is the point which is equidistant from all the vertices of ∆ABC

The point which is equidistant from all the sides of a triangle is called the in centre of the triangle.

In centre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle.

Here,

In ∆ABC, we can find the in centre of this triangle by drawing the angle bisectors of the interior angles of this triangle.

I is the point where these angle bisectors are intersecting each other.

Therefore,

I is the point equidistant from all

the sides of ∆ABC.

Maximum number of persons can approach the ice-cream parlour if it is equidistant from A, B and C.

Now,

A, B and C form a triangle.

In a triangle, the circum centre is the only point that is equidistant from its vertices.

So, the ice-cream parlour should be set up at the circum centre O of ∆ABC

In this situation,

Maximum number of persons can approach it.

We can find circum centre O of this triangle by drawing perpendicular bisectors of the sides of this triangle

It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles in it

Area of = (Side)²

= (5)²

= x 25

= cm²

Area of hexagonal shaped rangoli = 6 x

= cm²

Area of equilateral triangle having its side as 1 cm = (1)²

= cm²

Number of equilateral triangle of 1 cm side that can be filed in this hexagonal shaped rangoli = = 150

Star shaped rangoli has 12 equilateral triangles of side 5 cm in it

Area of star shaped rangoli = 12 * * (5)²

= 75

Number of equilateral triangles of 1 cm side that can b e filled in this star shaped rangoli =

= 300

Therefore, star shaped rangoli has more equilateral triangles in it