Question

In $\Delta ABC,E$ is mid-point of the median AD and BE produced meets side AC at point Q. Show that $BE:EQ=3:1$

Answer 1

Let's take point x on AC such that $BQ||DX$

$CD=DB$ (D is median)

Applyingmidpoint theorem in $\Delta CBQ$

$2DX=BQ$

Also $AE=ED$ (E is midpoint of AD)

Now applying midpoint theorem on $\Delta DAX$.

$2EQ=DX$

Hence $\frac{BE}{EQ}=\frac{BQ-EQ}{EQ}=\frac{2DX-\frac{DX}{2}}{\frac{DX}{2}}=\frac{3DX}{DX}=3:1$

Hence prooved.