# In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

(i) ΔPAC ~ ΔPDB (ii) PA.PB = PC. PD

**Solution:**

(i) In ΔPAC and ΔPDB

∠APC = ∠BPD (Common angle)

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

⇒ ΔPAC ~ ΔPDB

(ii) In ΔPAC and ΔPDB

PA/PD = PC/PB = AC/BD

PA/PD = PC/PB

PA.PB = PC.PD

**Video Solution:**

## In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) ΔPAC ~ ΔPDB (ii) PA.PB = PC.PD

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.6 Question 8:

In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) ΔPAC ~ ΔPDB (ii) PA.PB = PC.PD

In the above figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Hence prove that ΔPAC ~ ΔPDB and PA.PB = PC.PD.