Question

AP is the perpendicular bisector of BC​

Answer 1

Answer:

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In △s, ABD and ACD,

AD=AD (Common)

AB=AC (Given, ABC is an isosceles triangle)

DB=DC (Given, DBC is an isosceles triangle)

Thus, △ABD≅△ACD (SSS postulate)

Hence, PB=PC (By cpct)

and, ∠APB=∠APC=x (By cpct)

Now, ∠APB+∠APC=180

x+x=180

x=90°

Thus, AP is perpendicular bisector of BC

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