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Prove that the Circle Drawn on Any One of the Equal Sides of an Isosceles Triangle as Diameter Bisects the Base. - ICSE Class 10 - Mathematics

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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 – 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.

  Is there an error in this question or solution?

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Solution Prove that the Circle Drawn on Any One of the Equal Sides of an Isosceles Triangle as Diameter Bisects the Base. Concept: Arc and Chord Properties - Angle in a Semi-circle is a Right Angle.
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