In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG = AC.

From figure,

⇒ ∠FAG + ∠GAB + ∠BAD + ∠FAD = 360° [∵ At a point total angle is 360°]

⇒ ∠FAG + 90° + 90° + ∠BAD = 360°

⇒ ∠FAG = 360 – 90° – 90° – ∠BAD

⇒ ∠FAG = 180° – ∠BAD ………(i)

⇒ ∠ABC + ∠BAD = 180° [Sum of Adjacent angle in || gm is equal to 180°]

⇒ ∠ABC = 180° – ∠BAD ………(ii)

⇒ ∠FAG = ∠ABC [From (i) and (ii)]

In || gm ABCD,

AB = CD and AD = BC.

Since, ADEF is a square,

so, AD = DE = EF = FA.

So, BC = FA.

In ∆AFG and ∆ABC, we have

⇒ AF = BC [Proved above]

⇒ AG = AB (∵ AGBH is a square)

⇒ ∠FAG = ∠ABC [Proved above]

So, ∆AFG ≅ ∆ABC by S.A.S axiom of congruency

∴ FG = AC. [By C.P.C.T]

Hence, proved that FG = AC.