SSC CHSL Model Paper 14

Given: ABCD is a rhombus in which AB is produced to E and F such that
AE = AB = BF ....(1)
To prove: ED ⊥ FC
Proof:
Given ABCD is a rhombus.
Therefore, AB = BC = CD = DA ...(2)
On equating Eq. (1) and (2), we get:
BC = BF
⇒ ∠4 = ∠3 [angles opposite to equal sides are equal]
Again, ∠B (∠2) is the exterior angle of triangle BFC.
SO, ∠2 = ∠3 + ∠4 = 2 ∠4 ....(3)
Similarly, AE = AD, thereforce ∠5 = ∠6
Also ∠A (∠1)is the exterior angle of triangle ADE
So ∠1 = 2 ∠6 .... (4)
Also, ∠1 + ∠2 = 180° [consecutive interior angles]
Therefore, 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 ⊥ FG
Since ED and FC are part of EG and FG respectively, so ED is also perpendicular to FC.