Question

In parallelogram ABCD, E and F are mid point of the sides AB and CD respectively.The line segment AF and BF meet the line segment ED and EC at point G and H respectively prove that :
(i) Triangle HEB and FHC are congruent
(ii) GEHF is a parallelogram.

Answer 1

Given, $ABCD$ is a parallelogram. Since, $E$ and $F$ are midpoints of the sides $AB$ and $CD$ respectively,

$\therefore AE=BE=DF=CF$

$\left(1\right)$

Here, for $△HEB$ and $△FHC$,

$\angle EHB=\angle FHC[\because verticallyoppositeangles]\angle HFC=\angle HBE[AlternateinterioranglesofthelinesAB\parallel CDandBFistransversal]BE=CF\therefore △HEB\cong △FHC[byAngle-Angle-Sideproperty]$

$\left(2\right)$

In quadrilateral $AECF,$

$AE=CF$

and $AE\parallel CF(asAB\parallel CD)$

$\therefore AECFisaparallelogram[inaquadrilateral,ifapairofoppositesidesisequalandparallelthenitisaparallelogram]$

So, $EC\parallel AF\Rightarrow EH\parallel GF⟶\left(1\right)$

In quadrilateral $BEDF,$

$BE=DF$

and $BE\parallel DF(asAB\parallel CD)$

$\therefore BEDFisparallelogram$

So, $BF\parallel ED\Rightarrow HF\parallel GE⟶\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$GEHF$ is a parallelogram.