Question

In an isosceles AABC, D is the mid-point of BC. Prove that triangle ABD = triangle ACD. please solve this problem .PLEASE
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Answer 1

Step-by-step explanation:

I hope this will helps you

Answer 2

Answer:

Let DE and DF be perpendiculars from D on AB and AC respectively.

In order to prove that AB=AC, we will prove that ΔBDE≅ΔCDF.

In these two triangles, we have

∠BEF=∠CFD=90

∘

BD=CD [∵ D is the mid - point of BC ]

DE=DF [Given ]

So, by RHS congruence criterion, we obtain

ΔBDE≅ΔCDF

⇒∠B=∠C

⇒AC=AB

⇒ΔABC is isosceles