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 simpler

proof of this theorem in Class X.