Question
In BD and CE intersect each other at the point P. Is $\triangle\text{PBC}\sim\triangle\text{PDE}?$ Wht?
✓
Answer
True:If $\triangle\text{PBC}$ and $\triangle\text{PDE},$ we have
$\angle\text{BPC}=\angle\text{DPE}$ [Vertically opposite angles]
$\frac{\text{BP}}{\text{PD}}=\frac{5}{10}=\frac{1}{2}$
$\frac{\text{PC}}{\text{PE}}=\frac{6}{12}=\frac{1}{2}$
$\therefore\frac{\text{BP}}{\text{PD}}=\frac{\text{PC}}{\text{PE}}$
Hence, $\triangle\text{BPC}\sim\triangle\text{DPE}$ [by SAS similarity criterion]
Hence, the given statement is true.
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