# Ex. 6.6 Q8 Triangles Solution - NCERT Maths Class 10

## Question

In Fig. below, two chords \(AB\) and \(CD\) of a circle intersect each other at the point \(P\) (when produced) outside the circle.

**Prove that:**

(i) \(\Delta {\text{ }}PAC{\text{ }}\text{~}{\text{ }}\Delta {\text{ }}PDB\)

(ii) \(PA. PB = PC. PD \)

## Text Solution

**Reasoning:**

(i) Exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

**Steps:**

Draw \(AC\)

(i) In, \(\Delta PAC,\,\,\Delta PDB\)

\(\angle APC = \angle BPD\) (Common angle)

\(\angle PAC = \angle PDB\) (Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle)

\(\Rightarrow \Delta PAC\text{~}\Delta PDB\)

(iI) In, \(\Delta PAC,\,\,\Delta PDB\)

\[\begin{align} &\frac{{PA}}{{PD}} = \frac{{PC}}{{PB}} = \frac{{AC}}{{BD}}\\ &\frac{{PA}}{{PD}} = \frac{{PC}}{{PB}}\\ &PA \cdot PB = PC \cdot PD \end{align}\]