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 that ∆ABD ≅ ∆ACD.​

Answer 1

Answer:

hence proved

Step-by-step explanation:

in ΔABD and ΔACD

AB=AC (ΔABC is isosceles )

DB=DC ( ΔDBC is isosceles)

AD=AD (COMMON SIDE )

BY SSS congruency  ∆ABD ≅ ∆ACD.​

Answer 2

To Show :

1. ∆ABD ≅ ∆ACD.
2. ∆ABP ≅ △ACP
3. AP bisects ∠A as well as ∠D
4. AP is perpendicular bisector of BC .

Proof :

1. In △ABD AND △ACD

AB = AC [ GIVEN]

AD = AD [ COMMON ]

BD = CD [ GIVEN ]

Therefore, ∆ABD ≅ ∆ACD.[SSS]

2. In △ABP and △ ACP

AB = AC [ GIVEN ]

∠BAP = ∠CAP [ C.P.C.T ]

AP = AP [ COMMON ]

Therefore , △ABP ≅ △ACP [ SAS ]

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