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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.
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Solution
Given – In ∆ABC, AB = AC and a circle with AB as diameter is drawn
Which intersects the side BC and D.
To prove – D is the mid point of BC
Construction – Join AD.
Proof – ∠1 = 90° ...[Angle in a semi circle]
But ∠1 + ∠2 = 180° ...[Linear pair]
∴ ∠2 = 90°
Now in right ∆ABD and ∆ACD,
Hyp. AB = Hyp. AC ...[Given]
Side AD = AD ...[Common]
∴ By the right Angle – Hypotenuse – side criterion of congruence, we have
ΔABD ≅ ∆ACD ...[RHS criterion of congruence]
The corresponding parts of the congruent triangle are congruent.
∴ BD = DC ...[c.p.c.t]
Hence D is the mid point of BC.
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