Question

ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.

Answer 1

Given : ABCD is a rhombus, EABF is a straight line such that

EA = AB = BF

ED and FC are joined Which meet at G on producing

To prove : $\angle EGF={90}^{\circ }$

Proof : $\because$ Diagonals of a rhombus bisect each other at right angles

AO = OC, BO = OD

$\angle AOD=\angle COD={90}^{\circ }$

and $\angle AOB=\angle BOC={90}^{\circ }$

In $\Delta BDE,$

A and O are the mid-points of BE and BD respectively.

$\therefore$ AO || ED

Similarly , OC || DG

In $\Delta CFA,$ B and O are the mid-points of AF and AC respectively

$\therefore$OB || CF and OD || GC

Now in quad. DOCG,

OC || DG and OD || CG

$\therefore$ DOCG is a parallelogram.

$\therefore \angle DGC=\angle DOC$ (opposite angles of ||gm)

$\therefore \angle DGC={90}^{\circ }$ $(\because \angle DOC={90}^{\circ })$

Hence proved.