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