Question

"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 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.

Class 9 - Math - Areas of Parallelograms and Triangles Page 166"

Answer 1

Given:
ABC is a right angled triangle in which ∠A= 90°. BCED,ACFG,ABMN are squares.

To show:
WRITE FROM QUESTION

Proof:

(i) We know that each angle of a square is 90°.
Hence, ∠ABM = ∠DBC = 90º
∠ABM + ∠ABC = ∠DBC + ∠ABC
∠MBC = ∠ABD
In ΔMBC and ΔABD,
∠MBC = ∠ABD (Proved above)
MB = AB (Sides of square ABMN)
BC = BD (Sides of square BCED)

∴ ΔMBC ≅ ΔABD (SAS congruence rule)

(ii) We have
ΔMBC ≅ ΔABD
ar (ΔMBC) = ar (ΔABD) ... (1)
It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square BDEC)
BD || AX (Two lines perpendicular to same line are parallel to each other)

ΔABD & rectangle BYXD are on the same base BD and between the same parallels BD and AX.

ar(∆ABD)= ½ ar(BYXD)
ar(BYXD)=2ar(∆ABD)
ar(BYXD) = 2ar (ΔMBC)..........(2)
[Using equation (1)]

(iii) ΔMBC & square ABMN are lying on the same base MB and between same parallels MB and NC.

ar(∆MBC)=1/2ar(ABMN)

2 ar (ΔMBC) = ar (ABMN)

ar (BYXD) = ar (ABMN) [Using equation (2)] ... (3)

(iv) We know that each angle of a square is 90°.
∴ ∠FCA = ∠BCE = 90º
∠FCA + ∠ACB = ∠BCE + ∠ACB
∠FCB = ∠ACE

In ΔFCB and ΔACE,
∠FCB = ∠ACE
FC = AC (Sides of square ACFG)
CB = CE (Sides of square BCED)
ΔFCB ≅ ΔACE (SAS congruence rule)

(v) It is given that AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC)

Hence, CE || AX (Two lines perpendicular to the same line are parallel to each other)


ΔACE & rectangle CYXE are on the same base CE and between the same parallels CE and AX.

Ar(∆ACE)= 1/2ar(CYXE)

ar (CYXE) = 2 ar (ΔACE) ... (4)

We had proved that
∴ ΔFCB ≅ ΔACE
ar (ΔFCB) ≅ ar (ΔACE) ... (5)

On comparing equations (4) and (5),

ar (CYXE) = 2 ar (ΔFCB) ... (6)

(vi)
ΔFCB & square ACFG are lying on the same base CF and between the same parallels CF and BG.

Ar(∆FCB)= 1/2ar(ACFG)

ar (ACFG) = 2 ar (ΔFCB)

ar (ACFG) = ar (CYXE) [Using equation (6)] ... (7)


(vii) From the figure, it is evident that
ar (BCED) = ar (BYXD) + ar (CYXE)

ar (BCED) = ar (ABMN) + ar (ACFG) [Using equations (3) and (7)]

____________________________
Hope this will help you.

Answer 2

just shift the pages

quite difficult.