Question

ABCD IS A PARALLELOGRAM.THE CIRCLE THROUGH A,B,C INTERSECT CD AT E.PROVE THAT AD=AE

Answer 1

Given: ABCD is a parallelogram. A circle through A, B, C intersects CD produced at E.
To prove: AE = AD
Proof:
∠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 (1) and (2)
∠AED + ∠ABC = ∠ADE + ∠ADC
⇒∠AED = ∠ADE (using (3))
In ∆AED,
∠AED = ∠ADE
⇒AD = AE (equal sides have equal angles opposite to them)

Answer 2

Hello Mate!

Given : In ||gm ABCD, CD is produced till E to make quadrilateral cyclic.

To prove : AE = AD

Proof : Since quadrilateral ABCD is cyclic so sum of opposite angle is 180°.

< ABC + < AED = 180°

Where, < ADC = < ABC ( Opposite angle of ||gm are equal )

< ADC + < AED = 180°

Now, < ADC + < ADE = 180° ( Linear pair )

< ADC = 180° - < ADE

180° - < ADE + < AED = 180°

< AED = < ADE

Since side opposite to equal angles are equal therefore,

AD = AE.

Hence proved.

Hacw great future ahead!