Question

abc and dbc are two isosceles triangles 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 thati) abd acdii) abp acpiii) ap bisects as well as

Answer 1

Answer:

Step-by-step

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)

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

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90°

Hence, AP is the perpendicular bisector of BC.

Thanks