ABCD is a cyclic quadrilateral X Y and Z are the distance from b d ,bc and cd respectively, prove that BD upon X equals to BC upon Y + CD upon Z
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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
Step-by-step explanation:
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