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Question

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.


Solution

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
https://www.meritnation.com/img/userimages/mn_images/image/zzzz.png
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)


Mathematics
Concise Mathematics
Standard X

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