Question

In $\Delta ABC$, point D and E lies on the line AB and AC respectively as shown in the figure. Find the measure of $\angle AED$.

Answer 1

$\text{In}\Delta ABC,$

$\frac{AD}{DB}=\frac{4}{3}\text{and}\frac{AE}{EC}=\frac{4.8}{3.6}=\frac{4}{3}$

$\therefore \frac{AD}{DB}=\frac{AE}{EC}=\frac{4}{3}$

We know that in a triangle, if a line segment intersects two sides and divides them in the same ratio, then it will be parallel to the third side.

$\therefore$ DE || BC

$\text{So,}\angle AED=\angle ACB={40}^{\circ }$
$\text{(Corresponding angles are equal)}$