$A B C D$ is a rhombus and $A B$ is produced to $E$ and $F$ such that $A E = A B = B F$ . Prove that $E D$ and $F C$ are perpendicular to each other.

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Solution

Given : $A B C D$ is a rhombus AB produced to E and F such that $A E = A B = B F$ .
Construction : Join ED and CF and produce it to meet at G,
To : $E D ⊥ F C$ prove
proof : AB is produced to points E and F such that
$A E = A B = B F$ (i)
Also since ABCD is a rhombus
$A B = C D = B C = A D$ (ii)
Now in $Δ B C F, B C = B F$ [from (i) & (ii)]
$1! = 2!$
$3! = 1! + 2!$ [exterior angle]
$3! = 22!$ (iii)
Similarly , $A E = E D$
$5! = 6!$
$4! = 5! + 6! = 25!$ $5! + 2! + E! G F = 180^{o}$
$4! = 25!$ (iv) $E! G F = 90^{o}$
by adding (iii) and (iv)
$4! + 3! = 25! + 22!$ $∴ 4!$ and $3!$ are consective interior angles
$∴ E G ⊥ F C$ Now in $Δ E G F$ Hence it proved

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