Q: 8 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 meets BC at Y. Show that:
(iv)
Since, ar (||gm BYXD) = 2 .ar (MBC) ..........(i) [already proved in 2nd part]
and, ar (sq. ABMN) = 2. ar (MBC)............(ii)
[Since ABMN and AMBC are on the same base MB and between same parallels MB and NC]
From eq(i) and eq (ii), we get