How i find the best tricky way for solving the question of class 9th of Areas of Parallelograms and Triangles chapter of class 9th give me the best and easy method for this question no.8(3) of exercise 9.4 In Fig. 9.34, 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:(iii) ar(BYXD) = ar(ABMN)
∆MBC and parallelogram ABMN are lying on the same base MB and between same parallels MB and NC.
2 ar (∆MBC) = ar (ABMN)
ar (∆YXD) = ar (ABMN) [From equation (ii)] … (iii)