Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. show that

Asked by shweta mondal | 12th Jan, 2014, 09:39: AM

Expert Answer:

begin mathsize 16px style Ar left parenthesis increment APB right parenthesis equals 1 half cross times BP cross times AM
Ar left parenthesis increment CPD right parenthesis equals 1 half cross times DP cross times CN
Ar left parenthesis increment APB right parenthesis cross times Ar left parenthesis increment CPD right parenthesis
equals 1 half cross times BP cross times AM cross times 1 half cross times DP cross times CN
equals 1 fourth cross times BP cross times AM cross times DP cross times CN..... left parenthesis straight i right parenthesis
Similarly comma
Ar left parenthesis increment APD right parenthesis cross times Ar left parenthesis increment BPC right parenthesis
equals 1 fourth cross times BP cross times AM cross times DP cross times CN.... left parenthesis ii right parenthesis
From space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
Ar left parenthesis increment APB right parenthesis cross times Ar left parenthesis increment CPD right parenthesis equals Ar left parenthesis increment APD right parenthesis cross times Ar left parenthesis increment BPC right parenthesis end style

Answered by  | 30th Nov, 2017, 02:29: PM