Two chords AB and CD of a circle intersect at a point P outside the ci

1.	Two chords AB and CD of a circle intersect at a point P outside the circle. Prove that: (i) ∆ PAC ~ ∆ PDB (ii) PA. PB = PC.PD
Answer» Given : AB and CD are two chords To Prove: (a) ∆ PAC - ∆ PDB (b) PA. PB = PC.PD Proof: ∠ABD + ∠ACD = 180° …(1) (Opposite angles of a cyclic quadrilateral are supplementary) ∠PCA + ∠ACD = 180° …(2) (Linear Pair Angles ) Using (1) and (2), we get ∠ABD = ∠PCA ∠A = ∠A (Common) By AA similarity-criterion ∆ PAC - ∆ PDB When two triangles are similar, then the rations of the lengths of their corresponding sides are proportional. ∴ PA/PD = PC/PB ⇒ PA.PB = PC.PD

Two chords AB and CD of a circle intersect at a point P outside the circle. Prove that: (i) ∆ PAC ~ ∆ PDB (ii) PA. PB = PC.PD

Answer»

Given : AB and CD are two chords

To Prove:

(a) ∆ PAC - ∆ PDB

(b) PA. PB = PC.PD

Proof: ∠ABD + ∠ACD = 180° …(1) (Opposite angles of a cyclic quadrilateral are supplementary)

∠PCA + ∠ACD = 180° …(2) (Linear Pair Angles )

Using (1) and (2), we get

∠ABD = ∠PCA

∠A = ∠A (Common)

By AA similarity-criterion ∆ PAC - ∆ PDB

When two triangles are similar, then the rations of the lengths of their corresponding sides are proportional.

∴ PA/PD = PC/PB

⇒ PA.PB = PC.PD