In fig, 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) = 2ar(MBC) (iii)ar(BYXD) = ar(ABMN) (iv)△FCB≅△ACE (v)ar(CYXE) = 2ar(FCB) (vi)ar(CYXE) = ar(ACFG) (vii)ar(BCED) = ar(ABMN) + ar(ACFG) Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.

(i)In△s MBC and ABD, we have BC=BD [Sides of the square BCED] MB=AB [Sides of the square ABMN]∠MBC=∠ABD [Since Each=90°+∠ABC] Therefore by SAS criterion of congruence, we have △MBC≅△ABD
(ii)△ABD and square BYXD have the same base BD and are between the same parallels BD and AX. Therefore ar(ABD) = 1/2ar(BYXD) But △MBC≅△ABD [Proved in part (i)] ⇒ar(MBC)=ar(ABD) Therefore ar(MBC)=ar(ABD)=1/2ar(BYXD) ⇒ar(BYXD)=2ar(MBC)
(iii)Square ABMN and △MBChave the same base MB and are between same parallels MB and NAC. Therefore ar(MBC)=1/2ar(ABMN) ⇒ar(ABMN)=2ar(MBC)=ar(BYXD) [Using part (ii)]
(iv)In△s ACE and BCF, we have CE=BC [Sides of the square BCED] AC=CF [Sides of the square ACFG] and ∠ACE=∠BCF [Since Each=90°+∠BCA] Therefore by SAS criterion of congruence, △ACE≅△BCF
(v)△ACE and square CYXE have the same base CE and are between same parallels CE and AYX. Therefore ar(ACE)=1/2ar(CYXE) ⇒ar(FCB)=1/2ar(CYXE) [Since △ACE≅△BCF, part (iv)] ⇒ar(CYXE)=2ar(FCB)
(vi)Square ACFG and △BCF have the same base CF and are between same parallels CF and BAG. Therefore ar(BCF)=1/2ar(ACFG) ⇒1/2ar(CYXE)=1/2ar(ACFG) [Using part (v)] ⇒ar(CYXE)=ar(ACFG)
(vii)From part (iii) and (vi) we have ar(BYXD)=ar(ABMN) and ar(CYXE)=ar(ACFG) On adding we get ar(BYXD)+ar(CYXE)=ar(ABMN)+ar(ACFG) ar(BCED)=ar(ABMN)+ar(ACFG)