In 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) \(\triangle \)MBC \(\cong \) \(\triangle \)ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ax(ABMN)
(iv) \(\triangle \)FCB \(\cong \) \(\triangle \)ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)

We have a right \(\triangle \)ABC such that BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively. Line segment AX \(\perp \) DE is also drawn such that it meets BC at Y.


(i) \(\angle \)CBD = \(\angle \)MBA [Each90°]
\(\therefore \angle CBD + \angle ABC = \angle MBA + \angle ABC \)

(By adding \(\angle \)ABC on both sides)
\(\Rightarrow \) \(\angle \)ABD = \(\angle \)MBC

In \(\triangle \)ABD and \(\triangle \)MBC, we have
AB = MB [Sides of a square]
BD = BC
\(\angle \)ABD = \(\angle \)MBC [Proved above]
\(\therefore \) \(\triangle \)ABD \(\cong \) \(\triangle \)MBC [By SAS congruency]


(ii) Since parallelogram BYXD and \(\triangle \)ABD are on the same base BD and between the same parallels BD and AX.

\(\therefore \) ar(\(\triangle \) ABD) = \(\frac{1}{2} \) ar(||gm BYXD)

But \(\triangle \)ABD \(\cong \) \(\triangle \)MBC [From (i) part]
[Since, congruent triangles have equal areas.]
\(\therefore \) ar(\(\triangle \)MBC) = \(\frac{1}{2} \) ar(||gm BYXD)
\(\Rightarrow \) ar(||gm BYXD) = 2ar (\(\triangle \)MBC)


(iii) Since, ar(||gm BYXD) = 2ar(\(\triangle \)MBC) …(1) [From (ii) part]
and ar(\(\square \) ABMN) = 2ar(\(\triangle \)MBC) …(2)
[ABMN and AMBC are on the same base MB and between the same parallels MB and NC]

From (1) and (2), we have

ar(BYXD) = ar(ABMN) .



(iv) \(\angle \)FCA = \(\angle \)BCE (Each 90°)
\(\Rightarrow \) \(\angle \)FCA+ \(\angle \)ACB = \(\angle \)BCE+ \(\angle \)ACB
[By adding \(\triangle \)ACB on both sides]
\(\Rightarrow \) \(\angle \)FCB = \(\angle \)ACE
In \(\triangle \)FCB and \(\triangle \)ACE, we have
FC = AC [Sides of a square]
CB = CE [Sides of a square]
\(\angle \)FCB = \(\angle \)ACE [Proved above]
\(\Rightarrow \) \(\triangle \)FCB \(\cong \) \(\triangle \)ACE [By SAS congruency]


(v) Since, ||gm CYXE and \(\triangle \)ACE are on the same base CE and between the same parallels CE and AX.

\(\therefore \) ar(||gm CYXE) = 2ar((\(\triangle \)ACE)
But \(\triangle \)ACE \(\cong \) \(\triangle \)FCB [From (iv) part]
[Since, congruent triangles are equal in areas.]

\(\therefore \) ar (||gm CYXE) = 2ar(\(\triangle \)FCB)


(vi) Since, ar(||gm CYXE) = 2ar (\(\triangle \)FCB) …(3) [From (v) part]

Also (quad. ACFG) and \(\triangle \) FCB are on the same base FC and between the same parallels FC and BG.

\(\Rightarrow \) ar(quad. ACFG) = 2ar((\(\triangle \)FCB) …(4)

From (3) and (4), we get
ar( CYXE) = ar(ACFG) …(5)



(vii) We have ar(BCED) = ar(CYXE) + ar(BYXD)
= ar(CYXE) + ar(ABMN)
[From (iii) part]
Thus, ar (BCED) = ar( ABMN) + ar(ACFG)
[From (vi) part]