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Question is

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)    ΔMBC ≅ ΔABD
(ii)    ar(BYXD) = 2 ar(ΔMBC)
(iii) ar(BYXD) = ar(ΔABMN)
(iv)    ΔFCB ≅ ΔACE
(v)    ar(CYXE) = 2 ar(ΔFCB)
(vi)    ar(CYXE) = ar(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.

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