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A Parallelogram and a Triangle between the Same Parallels

In the figure...

Question

In the figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $A X ⊥ D E$ meets BC at Y Show that
$(i) △ M B C ≅ △ A B D (i i) a r (B Y X D) = 2 a r (△ M B C) (i i i) a r (B Y X D) = a r (A B M N) (i v) △ F C B ≅ △ A C E (v) a r (C Y X E) = 2 a r (△ F C B) (v i) a r (C Y X E) = a r (A C F G) (v i i) a r (B C E D) = a r (A M B N) + a r (A C F G)$

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Solution

Given : $I n △ A B C, ∠ A = 90^{\circ}$
BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively
$A X ⊥ D E$ meeting DE at X
To Prove:

(i) In $△ M B C a n d △ A B D, M B = A B B C = B D ∠ M B C = ∠ A B D ∴ M B C ≅ △ A B D (i i) ∵ △ A B D$ and rectangle BYXD are on eh same base BD and between the same parallels
$∴ a r (△ A B D) = \frac{1}{2} a r (r e c t . B Y X D) \Rightarrow a r (r e c t . B Y X D) = 2 a r (△ A B D) \Rightarrow a r (r e c t . B Y X D) = 2 a r (△ M B C)$
(iii) Similarly, $△ M B C$ and MBAN are on he same base MB and between the same parallels
$∴ a r (△ M B C) = a r (s q . A B M N)$
From (ii) and (iii)
ar(sq. ABMN ) = ar(rect. BYXD)
In $△ F C B a n d △ A C E, F C = A C C B = C E ∠ F C B ≅ ∠ A C E ∴ △ F C B ≅ △ A C E$
(v) $∵ △ F C B ≅ △ A C E ∴ a r (△ F C B) = a r (△ A C E) ∵ △ A C E$ and rectangle CYXE are on the same base and between he same parallels.
$∴ 2 a r (△ A C E) = a r (C Y X C) \Rightarrow 2 a r (△ F C B) = a r (C Y X E)$
(vi) $△ F C B$ and rectangle FCAG are on eh base FC and between the same parallels
$∴ 2 a r (△ F C B) = a r (F C A G)$
From (iv) and (vi)
$B C^{2} = A B^{2} + A C^{2} \Rightarrow B C \times B D = A B \times M B + A C \times F C \Rightarrow a r (B C E D) = a r (A B M N) + a r (A C F G)$

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Similar questions

Q. If the given figure, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.

Show that
(i) Δ MBC

≅

Δ ABD

(ii) ar (BYXD) = 2 ar (Δ MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) Δ FCB

≅

Δ ACE

(v) ar (CYXE) = 2 ar (ΔFCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)

Question 8
In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $A X ⊥ D E$ meets BC at Y. Show that:

(i) $Δ M B C ≅ Δ A B D$
(ii) $a r e a (B Y X D) = 2 a r e a (Δ M B C)$
(iii) $a r (B Y X D) = a r (A B M N)$
(iv) $Δ F C B ≅ Δ A C E$
(v) $a r (C Y X E) = 2 a r (Δ F C B)$
(vi) $a r (A C F G) = a r (C Y X E)$
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

Q. In fig,

A B C

is a right triangle right angled at

A . B C E D, A C F G

and

A B M N

are square on the sides

B C, C A

and

A B

respectively. Line segment

A X ⊥ D E

meets

B C

Y

. Show that:
(i)

△ M B C ≅ △ A B D

(ii)

a r (B Y X D) = 2 a r (M B C)

(iii)

a r (B Y X D) = a r (A B M N)

(iv)

△ F C B ≅ △ A C E

(v)

a r (C Y X E) = 2 a r (F C B)

(vi)

a r (C Y X E) = a r (A C F G)

(vii)

a r (B C E D) = a r (A B M N) + a r (A C F G)

Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler
proof of this theorem in Class X.

In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:

(i) ΔMBC ≅ ΔABD

(ii)

(iii)

(iv) ΔFCB ≅ ΔACE

(v)

(vi)

(vii)

Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in class X.

Question 8
In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $A X ⊥ D E$ meets BC at Y. Show that:

(i) $Δ M B C ≅ Δ A B D$
(ii) $a r e a (B Y X D) = 2 a r e a (Δ M B C)$
(iii) $a r (B Y X D) = a r (A B M N)$
(iv) $Δ F C B ≅ Δ A C E$
(v) $a r (C Y X E) = 2 a r (Δ F C B)$
(vi) $a r (A C F G) = a r (C Y X E)$
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

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