In Fig 6.4, BD and CE intersect each other at point P. Is △PBC ~ △PDE? Why

Solution:

Given, BD and CE intersect each other at point P.

We have to check if the triangles PBC and PDE are similar.

SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.

From the given figure,

∠BPC and ∠EPD are vertically opposite angles.

So, ∠BPC = ∠EPD

By the property of a similar triangle,

PB/PD = 5/10 = 1/2

PC/PE = 6/12 = 1/2

So, the sides are divided in the same ratio.

Since the two sides are in proportion and the included angles are congruent. The triangles PBC and PDE are similar triangles.

Therefore, △PBC ~ △PDE

✦ Try This: In the given figure, AD and CE intersect each other at point B. Is △DBC ~ △ADE? Why?

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6

NCERT Exemplar Class 10 Maths Exercise 6.2 Problem 4

In Fig 6.4, BD and CE intersect each other at the point P. Is △PBC ~ △PDE? Why

Summary:

In Fig 6.4, BD and CE intersect each other at the point P. △PBC ~ △PDE as two sides are in proportion and included angles are congruent.

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