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प्रश्न
In figure, BD and CE intersect each other at the point P. Is ΔPBC ~ ΔPDE? Why?
उत्तर
In ∆PBC and ∆PDE,
∠BPC = ∠EPD ...[Vertically opposite angles]
Now, `("PB")/("PD") = 5/10 = 1/2` ...(i)
And `("PC")/("PE") = 6/12 = 1/2` ...(ii)
From equations (i) and (ii),
`("PB")/("PD") = ("PC")/("PE")`
Since, one angle of ∆PBC is equal to one angle of ∆PDE and the sides including these angles are proportional, so both triangles are similar.
Hence, ∆PBC ~ ∆PDE, by SAS similarity criterion.
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