Question

Prove that circle drawn on any one of the equal side of an isosceles triangle as diameter bisect the base of the triangle

Answer 1

Given: ΔABC is an isosceles triangle with AB = AC. A circle is drawn taking AB as the diameter which intersects the side BC at D.

To prove: BD = DC

Construction: Join AD

Proof: ∠ADB = 90° (Angle in a semi-circle is 90°)

∠ADB + ∠ADC = 180°

⇒ ∠ADC = 90°

In ΔABD and ΔACD,

AB = AC   (Given)

∠ ADB = ∠ ADC (Proved)

AD = AD (Common)

⇒ ΔABD  ΔACD   (RHS congruence criterion)

⇒BD = DC (C.P.C.T)

Answer 2

HIII....

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