Question

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


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Answer 1

Answer:

From the figure,

⇒∠ADE+∠ADC=180

o

(Linear pair) ....... (i)

Also, □ABCE is a cyclic quadrilateral.

So, ∠AED+∠ABC=180

o

....... (ii) (opposite angles of cyclic quadrilateral sum upto 180

o

)

∠ADC=∠ABC ....... (iii) (Opposite angles of a parallellogram are equal)

Using (i), (ii) and (iii),

∠AED+∠ABC=∠ADE+∠ABC

⇒∠AED=∠ADE

Now, In △AED,

∠AED=∠ADE

⇒AD=DE (Sides opposite to equal angles)