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

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

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

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

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

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

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