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Question

ABCD is a parallelogram. The circle through A,B and C intersect CD (produced if necessary) at E. Prove that AE=AD.


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Solution

Step 1: Drawing the diagram:

A circle is drawn from the points A,B and C such that it intersects CD at E.

ABCD is a parallelogram

Step 2: Proving AE=AD:

From figure, we get,

ADE+ADC=180°…………(Linear pair)……(1)

As, ABCE is a cyclic quadrilateral.

AED+ABC=180°…………(Sum of the Opposite angles is 180°)………(2)

As, ABCD is a parallelogram

ADC=ABC……………………(Opposite angles of parallelogram are equal)………..(3)

Using, (1), (2) and (3), we get,

ADE+ADC=AED+ADC,

Therefore, ADE=AED

Now, In AED,

ADE=AED,

AE=AD…………….(Sides opposite to equal angles are equal).

Hence, proved.


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