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  • In Fig 6.4, BD and CE intersect each other at the point P. Is \DeltaPBC \sim \DeltaPDE? Why?

In Fig 6.4, BD and CE intersect each other at the point P. Is \DeltaPBC \sim \DeltaPDE? Why?

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Answer : [True]

Given:- BD and CE intersect each other at point P.

\text{Here, }  \angle BPC=\angle EPD \text{[Vertically opposite angle] }

\frac{BP}{PD}=\frac{5}{10}=\frac{1}{2}

\frac{PC}{PE}=\frac{6}{12}=\frac{1}{2}

\frac{BP}{PD}=\frac{PC}{PE}

Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.

\text{Also in } \Delta PBC \text{ and }\Delta PDE

\frac{BP}{PD}=\frac{PC}{PE}\; and\; \angle BPC=\angle EPD

\therefore \Delta PBC\sim \Delta PDE

Hence given statement is true.

Posted by

infoexpert23

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