In the given figure, squares ABDE and AFGC are drawn on the side AB and hypotenuse AC of right triangle ABC and BH ⟂ FG. Prove that :

(i) ∠EAC = ∠BAF

(ii) ar (sq. ABDE) = ar (rect. 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

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.