ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC. - Noon Academy

ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.

Circles | Class 10 | Exercise 17A | ICSE | Maths | Selina

ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(A) - 38

We assume that

$ABCD$

be the given cyclic quadrilateral.

$PA\text{ }=\text{ }PD\text{ }\left[ Given \right]$

So,

$\angle PAD\text{ }=\angle PDA\text{ }\ldots \ldots \text{ }\left( 1 \right)$

[Angles opposite to equal sides are equal]

And,

$\angle BAD\text{ }=\text{ }{{180}^{o}}-\angle PAD$

[Linear pair of angles]

Similarly,

$\angle CDA\text{ }=\text{ }{{180}^{o}}-\angle PDA$

$=\text{ }{{180}^{o}}-\angle PAD\text{ }\left[ From\text{ }\left( 1 \right) \right]$

As the opposite angles of a cyclic quadrilateral are supplementary,

$\angle ABC\text{ }=\text{ }{{180}^{o}}-\angle CDA$

$=\text{ }{{180}^{o}}-\text{ }({{180}^{o}}-\angle PAD)\text{ }=\angle PAD$

And,

$\angle DCB\text{ }=\text{ }{{180}^{o}}-\angle BAD$

$=\text{ }{{180}^{o}}-\text{ }({{180}^{o}}-\angle PAD)\text{ }=\angle PAD$

Hence,

$\angle ABC\text{ }=\angle DCB\text{ }=\angle PAD\text{ }=\angle PDA$

And it’s is only possible when

$AD\text{ }||\text{ }BC.$

i

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