Question

A quadrilateral ABCD is drawn to circumscribe a circle prove that AB + CD =AD+BC.


please answer me fast......

Answer 1

AP=AS(Tangents from a common point are of equal length)
DS=DR(Same reason as above)
CR=CQ(Same reason again)
BP=BQ(Same reason)
Now
lhs=AB +CD
=(AP+PB)+(CR+RD)
=AS+BQ+QC+SD
=(AS+SD)+(BQ+QC)
=AD+BC
=RHS
Hence proved
Hope it helps
:)

Answer 2

hey mate

here is your answer

Because tangents drawn to an external point from a circle are equal:

AH=AE(1)

BF=BE(2)

DH=DG(3)

CF=CG(4)

Adding (1),(2),(3),(4):

AH+BF+DH+CF=AE+BE+DG+CG

(AH+DH)+(BF+CF)=(AE+BE)+(DG+CG)

AD+BC=AB+CD

QED

hope it's helps

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