Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
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.
Video Tutorials For All Subjects
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