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 ✕ CE = BE ✕ 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 $A E \times C E=B E \times E D$