# In the given figure, two chords $AB$ and $CD$ intersect each other at the point $P$. Prove that (i) $∆APC \sim ∆DPB$.(ii) $AP \times PB = CP \times DP$."

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Given:

Two chords $AB$ and $CD$ intersect each other at the point $P$.

To do:

We have to prove that

(i) $∆APC \sim ∆DPB$.

(ii) $AP \times PB = CP \times DP$.

Solution:

(i) In $\triangle \mathrm{APC}$ and $\triangle \mathrm{BPD}$,

$\angle \mathrm{APC}=\angle \mathrm{DPB}$           (Vertically opposite angles)

$\angle \mathrm{PAC}=\angle \mathrm{PDB}$              (Angles on the same segment are equal)

Therefore, by AA similarity,

$\triangle \mathrm{APC} \sim \triangle \mathrm{DPB}$

Hence proved.

(ii) In $\triangle \mathrm{APC}$ and $\triangle \mathrm{BPD}$,

$\angle \mathrm{APC}=\angle \mathrm{DPB}$           (Vertically opposite angles)

$\angle \mathrm{PAC}=\angle \mathrm{PDB}$              (Angles on the same segment are equal)

Therefore, by AA similarity,

$\triangle \mathrm{APC} \sim \triangle \mathrm{DPB}$

This implies,

$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{CP}}{\mathrm{PB}}$

$\mathrm{AP} \times \mathrm{PB}=\mathrm{CD} \times \mathrm{PD}$

Hence proved.

Updated on 10-Oct-2022 13:21:55