Question 8In the following 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) area(BYXD)=2area(ΔMBC)(iii) ar(BYXD)=ar(ABMN)(iv) ΔFCB≅ΔACE(v) ar(CYXE)=2ar(ΔFCB)(vi) ar(ACFG)=ar(CYXE)(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

Question 8In the following figure, ABC is a right triangle right angle

1.	Question 8In the following 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) area(BYXD)=2area(ΔMBC)(iii) ar(BYXD)=ar(ABMN)(iv) ΔFCB≅ΔACE(v) ar(CYXE)=2ar(ΔFCB)(vi) ar(ACFG)=ar(CYXE)(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Answer» Question 8 In the following 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) $Δ M B C ≅ Δ A B D$ (ii) $a r e a (B Y X D) = 2 a r e a (Δ M B C)$ (iii) $a r (B Y X D) = a r (A B M N)$ (iv) $Δ F C B ≅ Δ A C E$ (v) $a r (C Y X E) = 2 a r (Δ F C B)$ (vi) $a r (A C F G) = a r (C Y X E)$ (vii) ar (BCED) = ar(ABMN) + ar(ACFG)