Question

ABC and A 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 that
(1) ABD=ACD
(2) ABP =ACP
AP bisects A as well as D.
AP is the perpendicular bisector of BC.​

Answer 1

Answer:

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.

Please mark my answer branliest this question take alot of effort