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Ex.6.3 Q8 Triangles Solution - NCERT Maths Class 10

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Question

\(E\) is a point on the side \(AD\) produced of a parallelogram \(ABCD\) and \(BE\) intersects \(CD\) at \(F.\) Show that \(\Delta ABE \sim \Delta CFB.\)

Diagram

 Video Solution
Triangles
Ex 6.3 | Question 8

Text Solution

Reasoning:

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This may be referred to as the \(AA\) similarity criterion for two triangles.

Steps:

In \(\Delta \rm{A B E}, \Delta \rm{C F B}\)

\(\angle B A E=\angle F C B\)

\(\text{(opposite angles of a parallelogram)}\)

 

\(\angle A E B=\angle F B C\) 

\(\begin{align} \begin{bmatrix} \because AE \,||\, BC \; \text{ and } \; EB \\ \text{ is a transversal alternate angle}\end{bmatrix} \end{align}\)

 

\(\Rightarrow\;\, \Delta ABE \sim \Delta C F E\)

\(\text{(AA Criterion)}\)