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 (AA criteria)

(ii) In ΔPAC and ΔPDB

PA/PD = PC/PB = AC/BD

PA/PD = PC/PB

PA.PB = PC.PD

☛ Check: NCERT Solutions for Class 10 Maths Chapter 6

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

Summary:

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.

☛ Related Questions:

• In Fig. 6.56, PS is the bisector of ∠QPR of ΔPQR. Prove that QS/SR = PQ/PR.
• In Fig. 6.57, D is a point on hypotenuse AC of ΔABC , such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that: (i) DM² = DN.MC (ii) DN² = DM.AN.
• In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that: AC² = AB² + BC² + 2BC.BD.
• In Fig. 6.59, ABC is a triangle in which ∠ABC less than 90° and AD ⊥ BC. Prove that: AC² = AB² + BC² - 2BC × BD.