One of the basic question from similarity chapter in which we have been asked to prove ΔPBC∼ΔPDE if BD and CE intersect each other at P.
Understanding ICSE Mathematics Avichal Publication Similarity chapter 13(i), question no 4
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In ΔPBC and ΔPDE , we have
∠BPC=∠DPE [Vertically opposite angles]
BP/PD=5/10=1/2
PC/PE=612=1/2
∴BP/PD=PC/PE
Hence , ΔBPC∼ΔDPE [By SAS similarity criterion]