Answer :

Let us consider the isosceles triangle ABC with side AB= AC.

i. Draw a circle with diameter AB and draw another circle with diameter AC. Thus, we get



ii. Join AH. We get,



As we can see in yellow circle,


AHB = 90°


( angle in a semicircle is right angle )


Similarly, in purple circle,


AHC = 90° ( angle in a semicircle is right angle )


Consider, Δ AHB and Δ AHC


AB = AC ( ABC is isosceles triangle)


AH = AH ( common side)


AHB = AHC 90°


Δ AHB Δ AHC (by RHS congruent rule)


Thus, BH = BC (by CPCT)


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