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In figure BD and CE intersect each other...

In figure BD and CE intersect each other at the point P. Is`DeltaPBC~DeltaPDE`? Why?

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True.
we have ` (PB)/(PD)= 5/10 = 1/2` …(1)
and `(PC)/(PE)= 6/12= 1/2` …(2)
`(PB)/(PD) = (PC)/(PE)` [ from (1) and (2) ] …(3)
Now in `trianglePBC and trianglePDE`
` angleBPC= angleDPE` ( vertically opposite angles)
and ` ( PB)/(PD) = (PC)/(PE) `
` trianglePBC ~ trianglePDE` (SAS similarity)
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