Question

In the given figure, $AD=DB$ and $AE=EC.$ The ratio, $BC:DE=$

Answer 1

It is given that, $AD=DB$
i.e, $AD=\frac{1}{2}AB$

Also, as $AE=EC$
we have, $AE=\frac{1}{2}AC$

By mid-point theorem, we know that the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is one half of the third side.
i.e, $DE\parallel BC$ and

DE = $\frac{1}{2}$ BC
Thus, $BC:DE=2:1$