here is the solution for your query:
given:
xy//ac.
ar(bxy)=ar(xyca)
to find:
ax/ab
proof:
in triangles abc and bxy,
xy//ac(given)
angle bxy=angle a and angle byx= angle c.(corresponding angles)
by aa similarity,
triangle abc similar to triangle bxy.
ar(abc)/ar(bxy)=(ab/bx)^2 (equation 1).
ar(abc)=2 ar(bxy) (given)
so,
ar(abc)/ar(bxy)=2/1 (equation 2)
from (1 and 2),
(ab/bx)^2=2/1
taking square root on both sides,
ab/bx=root 2/1
bx/ab=1/root 2.
1-bx/ab=1-1/root2
ab-bx/ab= root 2-1/root 2.
therefore,
ax/ab=root 2 -1/root 2=2-root 2/2.
hope it helped.
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