Ex. 6.6 Q8 Triangles Solution - NCERT Maths Class 10

Go back to  'Ex.6.6'


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



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


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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