In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.

If BH is perpendicular to FG, prove that :

(i) △ EAC ≅ △ BAF

(ii) Area of the square ABDE = Area of the rectangle ARHF.

(i) From figure,

⇒ ∠EAC = ∠EAB + ∠BAC

⇒ ∠EAC = 90° + ∠BAC (As, ABDE is a square and each angle of square equal to 90°) ……..(1)

Also,

⇒ ∠BAF = ∠FAC + ∠BAC

⇒ ∠BAF = 90° + ∠BAC (As, AFGC is a square and each angle of square equal to 90°) …………(2)

From equation (1) and (2),

⇒ ∠EAC = ∠BAF

In △ EAC and △ BAF,

⇒ EA = BA (Sides of square ABDE)

⇒ ∠EAC = ∠BAF (Proved above)

⇒ AC = AF (Sides of square AFGC)

∴ △ EAC ≅ △ BAF (By S.A.S. axiom)

Hence, proved that △ EAC ≅ △ BAF.

(ii) From figure,

ABC is a right angled triangle.

⇒ AC² = AB² + BC² [By pythagoras theorem]

⇒ AB² = AC² - BC²

⇒ AB² = (AR + RC)² - (BR² + RC²)

⇒ AB² = AR² + RC² + 2.AR.RC - BR² - RC²

⇒ AB² = AR² + RC² + 2.AR.RC - (AB² - AR²) - RC² [Using pythagoras theorem in △ ABR]

⇒ AB² = AR² + RC² + 2.AR.RC - AB² + AR² - RC²

⇒ AB² + AB² = AR² + AR² + RC² - RC² + 2.AR.RC

⇒ 2AB² = 2AR² + 2.AR.RC

⇒ 2AB² = 2AR(AR + RC)

⇒ AB² = AR(AR + RC)

⇒ AB² = AR.AC

⇒ AB² = AR.AF (As, AC = AF, sides of same sqaure)

∴ Area of square ABDE = Area of rectangle ARFH.

Hence, proved that area of square ABDE = area of rectangle ARFH.