In figure (2) given below, ABC is a right-angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that

(i) ∆BCF ≅ ∆ABE.

(ii) area of square ABFG = area of rectangle BENM.

(i) From figure,

⇒ ∠FBC = ∠FBA + ∠ABC

∠FBA = 90° (As each angle of a square = 90°)

⇒ ∠FBC = 90° + ∠ABC ……. (1)

Also,

⇒ ∠ABE = ∠CBE + ∠ABC

∠CBE = 90° (As each angle of a square = 90°)

⇒ ∠ABE = 90° + ∠ABC …….(2)

From (1) and (2), we get

∠FBC = ∠ABE …….. (3)

Now, in ∆BCF and ∆ABE

BF = AB (As FBAG is a square)

∠FBC = ∠ABE (Proved above)

BC = BE (As BCDE is a square)

By using the SAS axiom rule of congruency,

∴ ∆BCF ≅ ∆ ABE.

Hence, proved that ∆BCF ≅ ∆ ABE.

(ii) We know that,

∆BCF ≅ ∆ABE

So, area of ∆BCF = area of ∆ABE …… (4)

Since, ∆BCF and square AGFB have same base FB and are between same parallel lines FB and GC.

area of ∆BCF = $\dfrac{1}{2}$ area of square AGFB ……..(5)

∆ABE and rectangle BENM are on same base BE and are between same parallel lines BE and AN.

area of ∆ABE = $\dfrac{1}{2}$ area of rectangle BENM ……. (6)

From (4), (5) and (6)

⇒ $\dfrac{1}{2}$ area of square AGFB = $\dfrac{1}{2}$ area of rectangle BENM

⇒ area of square AGFB = area of rectangle BENM

Hence, proved that area of square AGFB = area of rectangle BENM.