In the 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)=2ar(△MBC)(iii)ar(BYXD)=ar(ABMN)(iv)△FCB≅△ACE(v)ar(CYXE)=2ar(△FCB)(vi)ar(CYXE)=ar(ACFG)(vii)ar(BCED)=ar(AMBN)+ar(ACFG)

In the figure, ABC is a right triangle right angled at A. BCED, ACFG a

1.	In the 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)=2ar(△MBC)(iii)ar(BYXD)=ar(ABMN)(iv)△FCB≅△ACE(v)ar(CYXE)=2ar(△FCB)(vi)ar(CYXE)=ar(ACFG)(vii)ar(BCED)=ar(AMBN)+ar(ACFG)
Answer» In the 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 $A X ⊥ D E$ meets BC at Y Show that $(i) △ M B C ≅ △ A B D (i i) a r (B Y X D) = 2 a r (△ M B C) (i i i) a r (B Y X D) = a r (A B M N) (i v) △ F C B ≅ △ A C E (v) a r (C Y X E) = 2 a r (△ F C B) (v i) a r (C Y X E) = a r (A C F G) (v i i) a r (B C E D) = a r (A M B N) + a r (A C F G)$