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 $AX\perp DE$ meets BC at Y. Show that:

(i) $\Delta MBC\cong \Delta ABD$
(ii) $area\left(BYXD\right)=2area\left(\Delta MBC\right)$
(iii) $ar\left(BYXD\right)=ar\left(ABMN\right)$
(iv) $\Delta FCB\cong \Delta ACE$
(v) $ar\left(CYXE\right)=2ar\left(\Delta FCB\right)$
(vi) $ar\left(ACFG\right)=ar\left(CYXE\right)$
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

Answer 1

(i) We know that each angle of a square is ${90}^{\circ }$.
Hence, $\angle ABM=\angle DBC={90}^{\circ }$
$\therefore \angle ABM+\angle ABC=\angle DBC+\angle ABC$
$\Rightarrow \angle MBC=\angle ABD$
In $\Delta MBCand\Delta ABD,$
$\angle MBC=\angle ABD\left(Provedabove\right)$
MB = AB (Sides of square ABMN)
BC = BD (Sides of square BCED)
$\Rightarrow \Delta MBC\cong \Delta ABD$ (SAS congruence rule)
$\left(ii\right)Wehave,\Delta MBC\cong \Delta ABD$
$\therefore ar\left(\Delta MBC\right)=ar\left(\Delta ABD\right)...\left(1\right)$
It is given that, $AX\perp DE$ and $BD\perp DE$. (Adjacent sides of square BDEC)
$\Rightarrow BD||AX$ (Two lines perpendicular to same line are parallel to each other)
$\Delta ABD$ and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.
$\therefore ar\left(\Delta ABD\right)=\frac{1}{2}ar\left(BYXD\right)$
i.e., $ar\left(BYXD\right)=2ar\left(\Delta ABD\right)$
$area\left(BYXD\right)=2area\left(\Delta MBC\right)$ [Using equation (1)] ... (2)
(iii) $\Delta MBC$ and parallelogram ABMN are lying on the same base MB and between same parallels MB and NC.
$\therefore ar\left(\Delta MBC\right)=\frac{1}{2}ar\left(ABMN\right)$
i.e., $2ar\left(\Delta MBC\right)=ar\left(ABMN\right)$
$\therefore ar\left(BYXD\right)=ar\left(ABMN\right)$ [Using equation (2)] ... (3)
(iv) We know that each angle of a square is ${90}^{\circ }$.
$\therefore \angle FCA=\angle BCE={90}^{\circ }$
Then, $\angle FCA+\angle ACB=\angle BCE+\angle ACB\left[Adding\angle ACBonbothsides\right]$
$\Rightarrow \angle FCB=\angle ACE$
In $\Delta FCBand\Delta ACE$,
$\angle FCB=\angle ACE$ [proved above]
FC = AC (Sides of square ACFG)
CB = CE (Sides of square BCED)
$\therefore \Delta FCB\cong \Delta ACE$ (SAS congruence rule)
(v) It is given that, $AX\perp DE$ and $CE\perp DE$. (Adjacent sides of square BDEC)
Hence, CE || AX (Two lines perpendicular to the same line are parallel to each other)
Consider $\Delta ACE$ and parallelogram CYXE.
$\Delta ACE$ and parallelogram CYXE are on the same base CE and between the same parallels CE and AX.
$\therefore ar\left(\Delta ACE\right)=\frac{1}{2}ar\left(CYXE\right)$
i.e., $ar\left(CYXE\right)=2ar\left(\Delta ACE\right)...\left(4\right)$
We had proved that,
$\Delta FCB\cong \Delta ACE$
$\therefore ar\left(\Delta FCB\right)=ar\left(\Delta ACE\right)...\left(5\right)$
On comparing equations (4) and (5), we obtain,
$ar\left(CYXE\right)=2ar\left(\Delta FCB\right)...\left(6\right)$
(vi) Consider $\Delta FCB$ and parallelogram ACFG.
$\Delta FCB$ and parallelogram ACFG are lying on the same base CF and between the same parallel CF and BG.
$\therefore ar\left(\Delta FCB\right)=\frac{1}{2}ar\left(ACFG\right)$
i,$ear\left(ACFG\right)=2ar\left(\Delta FCB\right)$
$\therefore ar\left(ACFG\right)=ar\left(CYXE\right)$ [Using equation(6)]... (7)
(vii) From the figure, it is evident that,
ar (BCED) = ar(BYXD) + ar(CYXE)
$\therefore$ ar (BCED) = ar(ABMN) + ar(ACFG) [Using equations (3) and (7)]