HKDSE MATHEMATICS

1. Angle AFC = Angle ADC = rt. angle. So AFDC is a cyclic quad. ( angle in the same segment equal).
So angle DFB = angle C (ext. angle of cyclic quad.)
Therefore, triangle DBF similar triangle ABC (AAA).
2. By Pythagoras thm.,
FC^2 = BC^2 - BF^2 = AC^2 - AF^2
so a^2 - BF^2 = b^2 - (c - BF)^2
a^2 - BF^2 = b^2 - c^2 + 2cBF - BF^2
so BF = (a^2 + c^2 - b^2)/2c
But by cosine rule, b^2 = a^2 + c^2 - 2ac cos B,
that is (a^2 + c^2 - b^2)/2c = a cos B, so BF = a cos B .............(1)
In the same way, BD = c cos B...............(2)
Since triangle DBF is similar to triangle ABC
FD/AC = BD/AB
FD/b = (c cos B)/c, so FD = b cos B ...............(3)
By similar figure, area of triangle FBD/area of triangle ABC = (BD/AB)^2
= [(c cos B)/c]^2 = cos^2 B.
The whole process is repeated for triangle AFE and triangle EDC, we get
area of triangle AFE/area of triangle ABC = cos^2 A and
area of triangle EDC/area of triangle ABC = cos^2 C.
So sum of area of the 3 triangle/area of triangle ABC = cos^2 A + cos^2 B + cos^2 C.
Therefore, area of triangle DEF = area of triangle ABC - sum of the area of the 3 triangles.
so area of triangle DEF/area of triangle ABC = 1 - cos^2 A - cos^2 B - cos^2 C.