A quadrilateral ABCD is drawn to circumscribe a circle prove that AB + CD =AD+BC.
please answer me fast......
Attachments:
Answers
Answered by
2
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
:)
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
:)
Anonymous286:
np✌✌
thnk u so much
no problem...
whoch chapter question is this
which class
yaa
i know
i was just confirming it
Answered by
0
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
✔️✔️✔️
Attachments:
Similar questions