Question

ABC is a triangle, D is a point on AB such that $AD=\frac{1}{4}AB$ and E is a point on AC such that $AE=\frac{1}{4}$ AC. Prove that DE = $\frac{1}{4}BC$

Answer 1

Given : In $\Delta ABC$, we have

D is a point on AB such that

$AD=\frac{1}{4}$ AB and E is a point on AC such that AE = $\frac{1}{4}AC$

DE is joined.

To prove : $DE=\frac{1}{4}BC$

Construction : Take P and Q the mid points of AB and AC and join them

Proof : In $\Delta ABC,$ we have

$\because$ P and Q are the mid -points of AB and AC

$\because PQ=\frac{1}{2}BC$

Similarly in $\Delta APQ,$ we have

D and E are mid -points of AP and AQ

$DE=\frac{1}{2}PQ$ [From above put $PQ=\frac{1}{2}\times BC$]

$=\frac{1}{2}\left(\frac{1}{2}BC\right)=\frac{1}{4}BC$

Hence $DE=\frac{1}{4}BC.$