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 ⊥ DE meets BC at Y. Show that
(i) ∆MBC = ∆ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ax(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ax(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)
Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler, proof of this theorem in Class X.
