Question

In the different circles, Harshita draw two chorós AB and CD which intersect each other at
the point P internally and externally. Prove that

(1) TriangleAPC » TriangleDPB
(1) AP. PB = CP.DP​

Answer 1

Step-by-step explanation:

angel 2 + angle 1 = 180 degree ( ABCD is a cyclic quad)---(1)

angle1+angle 2 =180 ( linear pair )------(2)

from (1) and (2)

angle 1 + angle2 = angle 2 + angle D ( angle 2 got cancelled

angle1 = angle D

i) In triangle PAC and Triangle PBD.

angle p = angle p ( common)

angle1 = angle D (proved above)

ii) triangle PAC is congruent to triangle PDB ( by AA similarity )

=> Cross multiply pa = pc , pd =pb

after cross multiplying we got (AC upon DB)

=> PA×PB = PD ×PC

=> hence , proved