In Fig. 15.91, 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) Δ MBC ≅ ABD (ii) ar(BYXD) =2ar(Δ MBC)

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

(iv) Δ FCB ≅ Δ ACE

(v) ar(CYXE) = 2ar(Δ FCB)

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

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

(i) In , we have

MB = AB

BC = BD

And,

∠ MBC = ∠ ABD (Therefore, ∠MBC and ∠ABC are obtained by adding ∠ABC to right angle)

So, by SAS congruence rule, we have

Area () = Area ( (1)

(ii) Clearly, and rectangle BYXD are on the same base BD and between the same parallels AX and BD

Therefore,

Area ( = Area of rectangle BYXD

Area of rectangle BYXD = 2 Area ()

Area of rectangle BYXD = 2 Area () (2)

[Therefore, Area ( = Area (] From (1)

(iii) Since,

and square MBAN are on the same base MB and between the same parallel MB and NC

Therefore,

2 Area () = Area of square MBAN (3)

From (2) and (3), we have

Area of square MBAN = Area of rectangle BXYD

(iv) In , we have

FC = AC

CB = CE

And,

∠FCB = ∠ACE (Therefore, ∠FCB and ∠ACE are obtained by adding ∠ACB to a right angle)

So, by SAS congruence rule, we have

(v) We have,

Area ( Area (

Clearly,

and rectangle CYXE are on the same base CE and between the same parallel CE and AX

Therefore,

2 Area ( = Area of rectangle CYXE

2 Area ( = Area of rectangle CYXE (4)

(vi) Clearly,

and rectangle FCAG are on the same base FC and between the same parallels FC and BG

Therefore,

2 Area ( = Area of rectangle FCAG (5)

From (4) and (5), we get

Area of rectangle CYXE = Area of rectangle ACFG

(vii) Applying Pythagoras theorem in , we have

BC² = AB² + AC²

BC * BD = AB * MB + AC * FC

Area of rectangle BCED = Area of rectangle ABMN + Area of rectangle ACFG