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Question
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:
- ΔEAC ≅ ΔBAF
- Area of the square ABDE
- Area of the rectangle ARHF.
Solution
(i) ∠EAC = ∠EAB + ∠BAC
∠EAC = 90° + ∠BAC ...(i)
∠BAF = ∠FAC + ∠BAC
∠BAF = 90° + ∠BAC ...(ii)
From (i) and (ii), we get
∠EAC = ∠BAF
In ΔEAC and ΔBAF, we have, EA = AB
∠EAC = ∠BAF and AC = AF
∴ ΔEAC ≅ ΔBAF ...(SAS axiom of congruency)
(ii) Since ΔABC is a right triangle, We have,
AC2 = AB2 + BC2 ...(Using pythagoras theorm in ΔABC)
⇒ AB2 = AC2 - BC2
⇒ AB2 = (AR + RC)2 - (BR2 + RC2) ...(Since AC = AR + RC and Using Pythagoras Theorem in ΔBRC)
⇒ AB2 = AR2 + 2AR × RC + RC2 - (BR2 + RC2) ...(Using the identity)
⇒ AB2 = AR2 + 2AR × RC + RC2 - (AB2 - AR2 + RC2) ...(Using Pythagoras Theorem in ΔABR)
⇒ 2AB2 = 2AR2 + 2AR × RC
⇒ AB2 = AR(AR + RC)
⇒ AB2 = AR × AC
⇒ AB2 = AR × AF
⇒ Area (`square`ABDE) = Area(rectangle ARHF).
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