Question

ABCD is a cyclic quadrilateral in which AB and CD when produced meet in E and EA=ED.Prove that AD || BC​

Answer 1

Answer:

Given ABCD is a cyclic quadrilateral in which EA = ED

(i) In △AED, we have

EA = ED [given]

⇒ ∠1 = ∠2

[angles opposite to equal sides are equal]

Also, ∠3 = ∠2 [external angle of cyclic quad. = interior opposites angle]

∠1 = ∠2

and ∠3 = ∠2

⇒ ∠1 = ∠3

These are correspondence angles.

∴ AD || BC

(ii) ∠1 = ∠4 [external angle of cyclic quad. = Interior opposite angle]

But ∠1 = ∠3 [proved above]

∠1 = ∠4

and ∠1 = ∠3

⇒ ∠3 = ∠4

In △BEC, we have

∠3 = ∠4 ⇒ BE = CE [side opposite to equal angles are equal]

Step-by-step explanation:

BTS ARMY

Answer 2

Given:-

• ABCD is a cyclic quadrilateral

• EA = ED

To prove:-

• AD || BC

Solution:-

Now, In ∆ EAD

→ EA = ED

Lets ∠EAD = x

Therefore, ∠EAD = ∠EDA = x [ ∵ Opposite angle to equal sides are equal ]

→ ∠BCD + ∠DBA = 180° [ ∵ Opposite angles are equal in a cyclic quadrilateral]

→ ∠BCD + x = 180° [ ∵ DAB = DAE = x]

→ ∠BCD = 180° - x

Similarly,

→ ∠ABC = 180° - x

→ ∠DAB + ∠ABC = x + 180° - x

→ ∠DAB + ∠ABC = 180° ----(1)

Similarly,

→ ∠BCD + ∠CDA = 180° ----(2)

★ From eq(1) and eq(2) -

→ The adjacent interior angles are supplementary,

Therefore, AB || BC.

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