Question

two chords AB and CD of a cricle intersect each other at the point P (when produced) outside the circle. prove that 1-trianglePACsimilar to trianglePDB ,2-PA.PB=PC.PD

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Given : Two chords AB and CD of a circle intersect each other at the point P

We have to prove that:

1) $\triangle PAC\sim \triangle PDB$

2) $PA\times PB=PC\times PD$

Proof:

1) Since, ABCD is a cyclic quadrilateral,

Thus, by the property of cyclic quadrilateral,

$\angle CAB + \angle CDB=180^{\circ}$ ---------(1)

But,  

$\angle PAB = 180^{\circ}$

$\implies \angle PAC + \angle CAB = 180^{\circ}$ ------(2)

Equation (1) - equation (2),

$\angle CDB - \angle PAC = 0$

$\angle CDB = \angle PAC$

$\implies \angle ADB = \angle PAC$

Similarly, we can prove,

$\angle ABD=\angle PCA$

Also,

$\angle APC = \angle BPD$    ( Reflexive )

By AAA similarity postulate,

$\triangle PAC\sim \triangle PDB$

Hence, proved.

2) Since, the corresponding sides of the similar triangles are in the same proportion,

⇒ $\frac{PA}{PC}=\frac{PD}{PB}$

⇒  $PA\times PB=PC\times PD$

Hence, proved.