In fig, ABC is a right triangle right angled at A.BCED,ACFG and ABMN are square 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 simplerproof of this theorem in Class X.

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(i) In Δs  MBC and ABD, we haveBC=BD [Sides of the square BCED]MB=AB [Sides of the square ABMN]∠MBC=∠ABD [Since Each =90º+∠ABC]Therefore by SAS criterion of congruence, we haveΔMBC≅δABD(ii) ΔABD and square BYXD have the same base BD and are between the same parallels BD and AX. Therefore ar(ABD)=21​ar(BYXD)But ΔMBC≅ΔABD  [Proved in part (i)]⇒ar(MBC)=ar(ABD)Therefore ar(MBC)=ar(ABD)=21​ar(BYXD)⇒ar(BYXD)=2ar(MBC).(iii) Square ABMN and ΔMBC have the same base MB and are between same parallels MB and NAC. Therefore ar(MBC)=21​ar(ABMN)⇒ar(ABMN)=2ar(MBC) =ar(BYXD) [Using part (ii)](iv) In ΔsACE and BCF, we have CE=BC [Sides of the square BCED]AC=CF [Sides of the square ACFG]and ∠ACE=∠BCF [Since Each=90º+∠BCA]Therefore by SAS criterion of congruence, ΔACE≅ΔBCF(v) ΔACE and square CYXE have the same base CE and are between same parallels CE and AYX. Therefore ar(ACE)=21​ar(CYXE)⇒ar(FCB)=21​ar(CYXE) [Since ΔACE≅ΔBCF, 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)=21​ar(ACFG)⇒21​ar(CYXE)=21​ar(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)

Question Text	In fig, $A BC$ is a right triangle right angled at $A . BCE D, A CFG$ and $A BMN$ are square on the sides $BC, C A$ and $A B$ respectively. Line segment $A X ⊥ D E$ meets $BC$ at $Y$ . Show that: (i) $△ MBC ≅ △ A B D$ (ii) $a r (B Y X D) = 2 a r (MBC)$ (iii) $a r (B Y X D) = a r (A BMN)$ (iv) $△ FCB ≅ △ A CE$ (v) $a r (C Y XE) = 2 a r (FCB)$ (vi) $a r (C Y XE) = a r (A CFG)$ (vii) $a r (BCE D) = a r (A BMN) + a r (A CFG)$ Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
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