pls answer
Asked by Pandeyo | 31st Aug, 2009, 10:22: PM
In EDC and
ABC
EDC =
ABC (given)
ECD is common.
By AA corollary
EDC
ABC
⇒ED/ AB = CD/ BC ..............(1)
Similarly, EDB
BFC
⇒ED/ FC = BD/BC ...............(2)
Adding (1) and (2)
ED/AB + ED/ FC = CD/BC + BD/BC
ED (1/AB + 1/ FC) = 1/BC (CD + BD)
Now on sunbstituting AB=x,ED=y,FC=z
y (1/x + 1/z) = 1/BC (BC)
y (1/x + 1/z) = 1
1/x + 1/z = 1/y
This is the required proof.
Answered by | 1st Sep, 2009, 09:12: AM
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