Question

Given triangle ABC with medians AE, BF, CD; FH parallel and equal in length to AE; BH and HE are drawn; FE extended meets BH in G. Which one of the following statements is not necessarily correct?

• A. AFHF is a parallelogram
• B. $\overline{HE}=\overline{HG}$
• C. $\overline{BH}=\overline{DC}$
• D. $\overline{FG}=\frac{3}{4}\overline{AB}$

Answer 1

The correct option is B $\overline{HE}=\overline{HG}$

R.E.F image

$FH=AE$ & $FH||AE$

Solution :-

it two lines are parallel

and equal then they will be in parallelogram

Here $FH=AE$ & $FH||AE$ (give)

So, $AFHE$ is a parallelogram

Now, $AF=HE$ & $AF||HE$

We look in $△ACD$ & $HBD$

$DC=BH$ (corresponding side)

$\therefore \left(C\right)$ is true.

Now

$FG=FE+EG$

$=\frac{1}{2}AB+\frac{1}{2}BD$ (By mid-point theorem in $△ABC$)

$=\frac{1}{2}AB+\frac{1}{2}\times \frac{1}{2}AB$ ($D$ is the mid point of $AB$)

$=\frac{3}{4}AB$

$\therefore$ only option $B$ left. So Answer will be $B$.