In a circle, two chords AB and CD intersect at a point P inside the circle. Prove that (a) ∆PAC ~ ∆PDB (b) PA. PB = PC. PD

Home » In a circle, two chords AB and CD intersect at a point P inside the circle. Prove that (a) ∆PAC ~ ∆PDB (b) PA. PB = PC. PD

In a circle, two chords AB and CD intersect at a point P inside the circle. Prove that (a) ∆PAC ~ ∆PDB (b) PA. PB = PC. PD

Answer:

Given,

AB and CD are two chords

In ∆ PAC and ∆ PDB

∠???????????? = ∠????????????

By ???????? ???????????????????????????????????????? ???????????????????????????????????? ∆???????????? ~????????????

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

∴ $\begin{array}{l} \frac{{PA}}{{PD}} = \frac{PC}{PB}\\ \end{array}$

PB = PC. PD

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