What is the simplest way for solving the question of class 9th of exercise 9.4 of math of question no.8(6) give me the best and simple way for solving this question in simple way 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:(vi) ar(CYXE) = ar(ACFG)
Consider BFCB and parallelogram ACFG
BFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.
∴ar (ACFG) = 2 ar (∆FCB)
∴ar (ACFG) = ar (CYXE) [From equation (vi)] … (vii)