In the given figure, AD = DB and AE = EC. Then, $B C : D E =$ ___.

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Solution

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 congruent to one half of the third side.
i.e., DE || BC and DE = $\frac{1}{2}$ BC
Thus, BC : DE = 2 :1

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