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प्रश्न
In Fig. ABCD is a cyclic quadrilateral. A circle passing through A and B meets AD and BC in the points E and F respectively. Prove that EF || DC.
उत्तर
In order to prove that EF || DC. It is sufficient to show that ∠2 = ∠3.
Since ABCD is a cyclic quadrilateral.
∴ ∠1 + ∠3 = 180° ...(i)
Similarly, in the cyclic quadrilateral ABFE, we have
∠1 + ∠2 = 180° ...(ii)
From (i) and (ii), we get
⇒ ∠1 + ∠3 = ∠1 + ∠2
⇒ ∠3 = ∠2
Hence, EF || DC.
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An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle, to prove the theorem complete the activity.
Given: ABCD is cyclic,
`square` is the exterior angle of ABCD
To prove: ∠DCE ≅ ∠BAD
Proof: `square` + ∠BCD = `square` .....[Angles in linear pair] (I)
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`square` + ∠BAD = `square` ......[Theorem of cyclic quadrilateral] (II)
By (I) and (II)
∠DCE + ∠BCD = `square` + ∠BAD
∠DCE ≅ ∠BAD
