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Question
Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution
Join AD.
AB is the diameter.
∴ ∠ADB = 90° ...(Angle in a semi-circle)
But, ∠ADB + ∠ADC = 180° ...(Linear pair)
`=>` ∠ADC = 90°
In ΔABD and ΔACD,
∠ADB = ∠ADC ...(Each 90°)
AB = AC ...(Given)
AD = AD ...(Common)
ΔABD ≅ ΔACD ...(RHS congruence criterion)
`=>` BD = DC ...(C.P.C.T)
Hence, the circle bisects base BC at D.
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