If the given figure, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.
Show that
(i) Δ MBC
Δ ABD
(ii) ar (BYXD) = 2 ar (Δ MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) Δ FCB
Δ ACE
(v) ar (CYXE) = 2 ar (ΔFCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)