**Question:**

In Figure 4, two triangles ABC and DBC are on the same base BC in which ∠A = ∠D = 90°. If CA and BD meet each other at E, show that AE âœ• CE = BE âœ• DE.

**Solution:**

Given that, there are two triangles *ABC* and *DBC *are on the same base *BC* in which

∠*A* = ∠*D* = 90°. If *CA* and *BD *meet each other at *E*, then we have to prove that *AE × CE = BE × DE*

The following figure is given

From the above figure, we can easily see that $\triangle A B E$ and $\triangle D C E$ are similar triangles, therefore we can use the property of similar triangle.

$\frac{A E}{D E}=\frac{B E}{C E}$

Hence $A E \times C E=B E \times E D$