ABCD is a rhombus and AB is produced to E and F such that AE=Ab=BF.Prove that ED and FC are perpendicular to each other.

**Given: **ABCD is a rhombus in which AB is produced to E and F such that AE = AB = BF.... (1)

**To prove:** ED ⊥ FC.

**Construction:** Join ED and FC such that they meet at G

**Proof:**

Given ABCD is a rhombus.

Therefore, AB = BC = CD = AD ... (2)

On equating (1) and (2), we get

BC = BF

⇒∠4 = ∠3 [Angles opposite to equal sides are equal]

Again, ∠B is the exterior angle of triangle BFC.

Therefore, ∠2 = ∠3 + ∠4 = 2∠4 ... (3)

Similarly, AE = AD ⇒∠5 = ∠6

Also, ∠A is the exterior angle of triangle ADE

⇒∠1 = 2∠6 ... (4)

Also, ∠1 +∠2 = 180° [consecutive interior angles]

∴ 2∠6 + 2∠4 = 180°

⇒ ∠6 + ∠4 = 90° ... (5)

Now, in triangle EGF, by angle sum property of triangle

∠6 + ∠4 + ∠G = 180°

⇒ 90°+ ∠G = 180° [using (5)]

⇒ ∠G = 90°

Thus, EG ⊥ FC.

Now, ED being a part of EG, so ED is also perpendicular to FC.

Hence, ED ⊥ FC.

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