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

Go back to  'Ex.6.6'

## 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}

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