Question

$\Delta ABCand\Delta DBC$ are two isosceles triangle on the same base BC and vertices A and D are on the same side of BC. if AD is extended to intersect BC at p, show that

AP is the perpendicular bisector of BC

Answer 1

Congruence of triangles:

Two ∆’s are congruent if sides and angles of a triangle are equal to the corresponding sides and angles of the other ∆.

In Congruent Triangles corresponding parts are always equal and we write it in short CPCT i e, corresponding parts of Congruent Triangles.

It is necessary to write a correspondence of vertices correctly for writing the congruence of triangles in symbolic form.

Criteria for congruence of triangles:

SAS( side angle side):

Two Triangles are congruent if two sides and the included angle of a triangle are equal to the two sides and included angle of the the other triangle.

SSS(side side side):

Three sides of One triangle are equal to the three sides of another triangle then the two Triangles are congruent.

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Solution;

Given,

ΔABC and ΔDBC are two isosceles triangles in which AB=AC & BD=DC.

To Prove:

(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

Proof:

(i) In ΔABD and ΔACD,

AD = AD (Common)

AB = AC (given)

BD = CD (given)

Therefore, ΔABD ≅ ΔACD (by SSS congruence rule)

∠BAD = ∠CAD (CPCT)

∠BAP = ∠CAP

(ii) In ΔABP & ΔACP,

AP = AP (Common)

∠BAP = ∠CAP

(Proved above)

AB = AC (given)

Therefore,

ΔABP ≅ ΔACP

(by SAS congruence rule).

(iii)

∠BAD = ∠CAD (proved in part i)

Hence, AP bisects ∠A.

also,

In ΔBPD and ΔCPD,

PD = PD (Common)

BD = CD (given)

BP = CP (ΔABP ≅ ΔACP so by CPCT.)

Therefore, ΔBPD ≅ ΔCPD (by SSS congruence rule.)

Thus,

∠BDP = ∠CDP( by CPCT.)

Hence, we can say that AP bisects ∠A as well as ∠D.

(iv)

∠BPD = ∠CPD

(by CPCT as ΔBPD ≅ ΔCPD)

& BP = CP (CPCT)

also,

∠BPD + ∠CPD = 180° (BC is a straight line.)

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90°

Hence,

AP is the perpendicular bisector of BC.