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# 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]