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

SinceΔEAD=ΔABC
(Corresponding angles),&AD=BC(sides of the rhombus)
By side angle side congrugency,the triangles
Therefore the corresponding parts of the triangle are also equal.
ΔCBF & ΔDAB
Since the sides 
BF=AB
∠CBF∠DAB(corresponding angles)
ΔCBF&ΔDAB are congrugent
ΔAOB
∠ABO+∠BAO+∠AOB=180°
∠ABO+∠BAO=90°
∠CFB+∠AED=90°
∠XFE+∠XEF=90°

Answer 2

Answer:

First we have to prove ∆DEA CONGRUENT∆CBF than in ∆GEF angle E = angle F by congruence of two triangles above let both angle be = X than use angle sum property of triangle so angle G +angle E+ angle F=180 degree so ans comes angle G= 90