Question

वर्तुळाला ABCD हा चक्रिय चौकोन आहे तर AB+CD=AD+BC असे सिध्द करा.
C
R
D
Q
A
P
B​

Answer 1

Answer:

Given c(o,r) proof- let AB touches the circle at P, BC at Q, DC at R and AD at S.

then PB= PQB(length of tangents drawn from an external point are always equal)

QC=RC

AP=AS

DS=DP

Now, AB+CD=AP+PB+DR+RC=AS+QB+DS+CQ=AS+DS+QB+CQ=AD+BC

hence proved

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