In figure, BD and CE intersect each other at the point P. Is ∆PBC ∆PDE?

Why?



(True)

In ∆PBC and ∆PDE,


BPC = EPD [vertically opposite angles]



Since, BPC of PBC is equal to EPD of PDE and the sides including these.


So,


∆PBC ∆PDE by SAS similarity criteria.


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