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In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that:
(i) ΔAPC ~ ΔDPB (ii) AP.PB = CP.DP
Solution:
As we know that, two triangles, are similar if

Their corresponding angles are equal and

Their corresponding sides are in the same ratio
As we know that angles in the same segment of a circle are equal.
(i) Draw BC
In ΔAPC and ΔDPB
∠APC = ∠DPB (Vertically opposite angles)
∠PAC = ∠PDB (Angles in the same segment)
⇒ ΔAPC ~ ΔDPB (AA criterion)
(ii) In ΔAPC and ΔDPB,
AP/DP = CP/PB = AC/DB [∵ ΔAPC ~ ΔDPB]
AP/DP = CP/PB
⇒ AP.PB = CP.DP
☛ Check: NCERT Solutions for Class 10 Maths Chapter 6
Video Solution:
In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that: (i) ΔAPC ~ ΔDPB (ii) AP.PB = CP.DP
NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.6 Question 7
Summary:
In the above figure, two chords AB and CD intersect each other at point P. Hence it is proved that ΔAPC ~ ΔDPB and AP.PB = CP.DP
☛ Related Questions:
 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 Fig. 6.63, D is a point on side BC of ∆ABC such that BD/CD = BA/CA. Prove that: AD is the bisector of ∠BAC.
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