Question

In parallelogram $ABCD$, $E$ and $F$ are midpoints of the sides $AB$ and $CD$ respectively. The line segments $AF$ and $BF$ meet the line segments $ED$ and $EC$ at points $G$ and $H$ respectively.
Prove that:
(i) triangles $HEB$ and $FHC$ are congruent
(ii) $GEHF$ is a parallelogram

Answer 1

$\left(i\right)$Since $ABCD$ is a parallelogram, then

$AB=CD$ and $AD=BC$, as opposite sides of parallelogram are equal.

Now,$AB=CD$

Since,$E$ is mid point of $AB$ and $F$ is the mid point of $CD$, then

$AE=BE=CF=DF$

In $\Delta HBE$ and $\Delta HFC$

$\angle EHB=\angle FHC($ Vertically opposite angles$)$

$\angle HBE=\angle HFC($ Alternate interior angles as $AB||CD$ and $BF$ is a transversal $)$

$BE=CF$ (Proved above)

$\Delta HBE\cong \Delta HFC\left(AAS\right)$

$\left(ii\right)$In Quadrilateral $AECF,$

$AE||CF($ as $AB||DC)$

$\to AECF$ is a parallelogram $[$ in quad, if a pair of opposite sides is equal and parallel, then it is a parallelogram$]$

$\to EC||AF$ or $EH||GF...........\left(1\right)$

In Quadrilateral $BFDE$,

$BE||DF($ proved above$)$

$BE||DF($ as $AB||DC)$

$\to BEDF$ is a parallelogram $[$ in quad, if a pair of opposite sides is equal and parallel, then it is a parallelogram$]$

$\to BF||ED$ or $HF||EG.....\left(2\right)$

from (1) and (2), we get

$HFGE$ is a parallelogram.