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"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

Answered by nikitasingh79
2

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....

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