Question

ABCD is a parallelogram. E and F are mid-points of the sides AB and CO respectively. The straight lines AF and BF meet the straight lines ED and EC in the points G and H respectively. Prove that: 1) Triangle HEB= Triangle HCF. 2)GEHF is a parallelogram.

Answer 1

Answer:

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

∴AE=BE=DF=CF

(1)

Here, for △HEB and △FHC,

∠EHB=∠FHC[∵verticallyoppositeangles]

∠HFC=∠HBE[AlternateinterioranglesofthelinesAB∥CDandBFistransversal]

BE=CF

∴△HEB≅△FHC[byAngle−Angle−Sideproperty]

(2)

In quadrilateral AECF,

AE=CF

and AE∥CF(asAB∥CD)

∴AECFisaparallelogram[inaquadrilateral,ifapairofoppositesidesisequalandparallelthenitisaparallelogram]

So, EC∥AF

⇒EH∥GF⟶(1)

In quadrilateral BEDF,

BE=DF

and BE∥DF(asAB∥CD)

∴BEDFisparallelogram

So, BF∥ED

⇒HF∥GE⟶(2)

From (1) and (2) we get,

GEHF is a parallelogram.