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:
SInce ||gm BYXD and ABD are on the same base BD and between same parallels BD and AX
Therefore, ar(ABD) = 1/2. ar(||gm BYXD)..........(i)
But, ABD MBC (proved in 1st part)
Since congruent triangles have equal areas.
Therefore, By using equation (i) we get