Prove that

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)