Question

In given Fig., two chords AB and CD intersect each other at the point P Prove that:

(i) $\Delta \text{APC~}\Delta \text{DPB}$
(ii) AP . PB = CP . DP

Answer 1

(i)
Step 1: First identify the vertically opposite angle
Gien: In $\Delta \text{APC and}\Delta \text{DPB,}$
$\angle \text{APC}=\Delta \text{DPB}$ ...(i) [Vertically Opposite Angles]

Step 2: Identify angles in the same segment
Given: In $\Delta \text{APC and}\Delta \text{DPB,}$
$\angle \text{CAP}=\angle \text{BDP}$ ...(ii)
[Angles in same segment]

Step 3: Applying similarity in $\Delta \text{APC and}\Delta \text{DPB}$
Given: In $\Delta \text{APC and}\Delta \text{DPB,}$
$\angle \text{APC}=\angle \text{DPB}$ ...from (i)
$\angle \text{CAP}=\angle \text{BDP}$ ...from (ii)
$\therefore$ By AA - criterion of similarity,

$\Rightarrow \Delta \text{APC~}\Delta \text{DPB}$
Hence proved.

(ii)
Step 1: Identifying the vertically opposite angle
Given: In $\Delta \text{APC and}\Delta \text{DPB,}$
$\angle \text{APC}=\Delta \text{DPB}$ ...(i) [Vertically Opposite Angles]

Step 2: Identifying angles in the same segment
Given: In $\Delta \text{APC and}\Delta \text{DPB,}$
$\angle \text{CAP}=\angle \text{BDP}$ ...(ii)
[Angles in same segment]

Step 3: Applying similarity in $\Delta \text{APC and}\Delta \text{DPB}$
Given: In $\Delta \text{APC andd}\Delta \text{DPB},$
$\angle \text{APC}-\angle \text{DPB}$ ...from (i)
$\angle \text{CAP}=\angle \text{BDP}$ ...from (ii)
$\therefore$ By AA-criterion of similarity,

$\Rightarrow \Delta \text{APC~}\Delta \text{DPB}$
Hence proved.

Step 4: Using property of similar triangles.
In $\Delta \text{APC~}\Delta \text{DPB}$
Sides are proportional in similar triangles,
$\frac{\text{AP}}{\text{DP}}=\frac{\text{PC}}{\text{PB}}=\frac{\text{CA}}{\text{BD}}\Rightarrow \frac{\text{AP}}{\text{DP}}=\frac{\text{PC}}{\text{PB}}$

$\therefore \text{AP}\times \text{PB}=\text{CP}\times \text{DP}$
Hence proved.