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In ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that   \({AP\over PD} ={ PC\over BP}\)   

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Sol.       seg AD || seg BC
             Taking AC as the transversal.
            ∠DAC ≅ ∠BCA                                      [Converse of alternate angle test]
i.e        ∠DAP ≅ ∠BCP                                       ...(I)
            Consider ΔAPD and ΔCPB
            ∠DAP ≅ ∠BCP                                      [From (I)]
            ∠APD ≅ ∠CPB                                      [Vertically opposite angles]
∴          ΔAPD ~ ΔCPB                                      [AA test of similarity]

∴         \({AP\over CP} ={ PD\over BP}\)                                               [c.s.s.t]

∴         \({AP\over PD} ={ PC\over BP}\)                                               [By Alternendo]

 

Similarity August 22 , 2018 0 Comments 13 views