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Question

In the following figures, ABCD is a parallelogram. A circle through A, B, C intersects CD or CD produced at E. Prove that AE = AD.

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Solution




Given: A parallelogram ABCD and a circle through A, B and C.To prove: AE=ADConstruction: Join AEProof:From figure i;As, ABC=ADC Opposite angles of parallelogram ABCDBut ABC+AEC=180° Opposite angles of cyclic quadrilateral AECBSo, ADC+AEC=180°ADC+AED=180° .....1Also, ADC+ADE=180° Linear pair .....2From 1 and 2, we getADC+AED=ADC+ADEAED=ADE AD=AE Sides opposite to equal angles are equalSimilalry, from figure ii;As, ABC=ADC Opposite angles of parallelogram ABCDBut ABC+AEC=180° Opposite angles of cyclic quadrilateral AECBSo, ADC+AEC=180°ADE+AEC=180° .....1Also, AEC+AED=180° Linear pair .....2From 1 and 2, we getADE+AEC=AEC+AEDADE=AED AD=AE Sides opposite to equal angles are equal

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