Question

∆ABC is an isosceles triangle with AB = AC. AM is the bisector of
BAC. Prove that ∆ABM~= ∆ACM.
[Hint: Use ASA congruence criterion as BAM - CAM
AB = AC and ABC = ACM]
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Answer 1

Answer:

We have to draw AP perpendicular to BC. Since △ABC is isosceles, AB=AC.

Given, AP perpendicular to BC

∴∠APB=∠APC=90o

In △ABP and △ACP, 

∠APB=∠APC=90o

AP=AC;AP=AP [common]

△ABP≅△ACP [RHS Congruence Rule]

∴∠B=∠C [by CPCT]