Question

2. In
he adioining figure, ABC is a right triangle right
That A BCED, ACFG and ABMN are squares on
angle
the sides BC, CA and AB respectively. Line segment
AX 1 DE meets BC at Y.
Show that:
(1) AMBC = AABD
(ii) ar (rect. BYXD) = 2 ar (AMBC)
(ii) ar (rect. BYXD) = ar(square ABMN)
(iv) AFCB =AACE
(u) ar (rect. CYXE) = 2 ar (AFCB)
(vi) ar (rect. CYXE) = ar(sq. ACFG)
(vii) ar (sq. BCED) = ar (sq. ABMN) + ar (sq. ACFG).​

Answer 1

just shift the pages

quite difficult