Question

Triangle ABC and triangle DBC are two isosceles triangle on the same base BC If AD is extended to intersect BC at P show that

1) triangle ABD congruence triangle ACD

2) triangle ABC congruence triangle ACP

3) AP is bisects angle A as well as angle D

4) AP is the perpendicular bisector of BC

Answer 1

$\huge\textbf{Solution}$
$<b><i>$
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)


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

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90°

Hence, AP is the perpendicular bisector of BC.


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Answer 2

Answer:

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