Question

In a qudrilateral ABCD Prove that AB+CD+AD>BC​

Answer 1

Step-by-step explanation:

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+C

Q=AS+DS+QB+CQ=AD+BC

____________________

hence proved