"Question 6 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Class 9 - Math - Circles Page 186"
Answers
A quadrilateral is called a cyclic quadrilateral if all the four vertices of a quadrilateral lie on a circle are ( concyclic).
The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
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Given: ABCD is a parallelogram. A circle through A, B, C intersects CD produced at E.
To prove: AE = AD
Proof:
Since ABCE is a cyclic quadrilateral.
∠AED + ∠ABC = 180° ... (1)
(Sum of opposite angles of cyclic quadrilaterals is 180°)
∠ADE + ∠ADC = 180° ... (2)
(linear pair)
∠ABC = ∠ADC ... (3)
(opposite angles of parallelogram are equal)
From eq (1) and (2)
∠AED + ∠ABC = ∠ADE + ∠ADC
∠AED = ∠ADE
(from eq 3)
In ∆AED,
∠AED = ∠ADE
AD = AE
(since, sides opposite to equal angles of a triangle are equal).
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Hope this will help you....
