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In a qudrilateral ABCD Prove that AB+CD+AD>BC |
Answer» 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 |
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