ABCD is a square of side a. E is the midpoint of DA. F is a point other than D on DB such that
EFC
= 90o.
Question:
ABCD is a square of side a. E is the midpoint of
DA. F is a point other than D on DB such that EFC
= 90o.
To show : DF = 3 FB
EFC
= ADC
= 90o, we can show that DEFC is a cyclic quadrilateral. Making
use of the property of ‘angles in the same segment’ and the fact that CDF
= 45o, we can conclude that CEF
= 45o. Using the conditions for isosceles triangles, we can
show that FC = FE. By joining FA and using symmetry, we can deduce that
AF = FC. Hence AF = FE. Using the perpendicular bisector theorem, we can
show that F lies on the perpendicular bisector of EA. Then the ratio of
the lengths of DF and FB can be found easily by considering the lengths
of DG and GA.
(II) Using coordinate geometry
Without loss
of generality, let A, B, C and D be (a, a), (a, 0), (0, 0) and (0, a) respectively.
Then E will be (a/2, a). We can make use of the fact that EFC
= 90o to find out the locus of F. This can be found by considering
the slopes of EF and FC. Then we can get an equation relating the x-coordinate
and y-coordinate of F. Hence, by simplifying, the ratio of the lengths
of DF and FB could be obtained.
(III) Using vectors
Without loss
of generality, let C be the origin, CD = b and CB
= c. Then by putting BF = d BD, we want to show that
d = 1/4. Obviously, we can express CE and CF in terms of
b and c. Hence, we can express FE in terms of b
and c. The condition that EFC
= 90o is equivalent to that the dot product of FE and
CF is zero. After some simplications, d can be solved and hence
the required ratio.
(IV) Using complex numbers
Without loss
of generality, let the complex number 0 be represented by B in the Argand
diagram. Let x, y and z be the complex numbers corresponding to C, E and
F respectively, and let e be the length of FB. Obviously, we can find out
x, y and z by considering their modulus and argument as follows:
x = - a
y = - (a/2) + a i
z = e [cos(3p /4) + i sin(3p
/4)]
As EFC
= 90o, CE2
= EF2 + CF2
. Then we have
| y - x |2
= | z - x |2
+ | z - y |2
After some simplications, we can get an equation involving a and e. Solving this quadratic equation, we can express e in terms of a. Hence, we can find out the lengths of BF and FD and therefore their ratio.
(V) Using the properties of cyclic quadrilaterals and the sine rule
Let BCF
= g. We can get the following by using the properties of cyclic quadrilaterals:
DCF
= 90o - g , DFC
= 45o + g , EFD
= 45o - g
and DEF
= 90o + g.
Using the sine rule in DCDF, we get
and from DDEF, we get 
From these three sets of equations, we can show that the ratio of DF to FB is cot g : 1. It suffices to show from these equations that cot g = 3 or tan g = 1/3.
(VI) Using the condition that FC = FE and the cosine rule
The condition that FC = FE can be proved by using the properties of cyclic quadrilaterals and the conditions for isosceles triangles.
By drawing
the perpendicular from E onto CB (such that EG 
CB) and using the condition that FC = FE, we can find the length of EC
in terms of a and hence the length of FC. Using the cosine rule for
DCDF, we get
FC2 = CD2
+ DF2 - 2(CD)(DF)
cos 45o
and similarly for DBCF, we get
FC2 = BC2
+ FB2 - 2(BC)(FB)
cos 45o
Then, we can observe that DF and FB are the two roots
of a quadratic equation. Solving this quadratic equation, we get
DF = 3 FB.
(VII) Combining the proofs of (V) and (VI)
We can combine
different parts of the proofs in (V) and (VI). The main difference
is that we can find the exact measure of the angles concerned. Firstly,
we can express FC in terms of a. Secondly, we can find the exact
magnitude of CEG.
Hence we can find FEG
and BCF.
Then, based on the length of FC and by considering DHCF
and DHBF, we can express FH and FB in terms
of a. As we can easily compute the length of DB, the required ratio
can be solved.
