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$ABCD$ is a rhombus, $EABF$ is a straight line such that $EA = AB = BF$. Prove that $ED$ and $FC$ when produced meet at right angles.
Given:
$ABCD$ is a rhombus, $EABF$ is a straight line such that $EA = AB = BF$.
To do:
We have to prove that $ED$ and $FC$ when produced meet at right angles.
Solution:
Let $ED$ and $FC$ when joined meet at $G$ on producing.
We know that,
Diagonals of a rhombus bisect each other at right angles.
This implies,
$\angle AOD = \angle COD =\angle AOB =\angle BOC = 90^o$
$AO = OC, BO = OD$
In $\triangle BDE$,
$A$ and $O$ are the mid-points of $BE$ and $BD$ respectively.
This implies,
$AO \parallel ED$
Similarly,
$OC \parallel DG$
In $\triangle CFA, B$ and $O$ are the mid-points of $AF$ and $AC$ respectively.
Therefore,
$OB \parallel CF$ and $OD \parallel GC$
In quadrilateral $DOCG$,
$OC \parallel DG$ and $OD \parallel CG$
This implies,
$DOCG$ is a parallelogram.
Therefore,
$\angle DGC = \angle DOC$ (Opposite angles of a parallelogram are equal)
This implies,
$\angle DGC = 90^o$
Hence proved.
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