Austin draws \(\triangle ABC\) with \(AB=3\) cm, \(BC=4\) cm, and \(\angle ABC=90\degree\). Lachlan then draws \(\triangle DBF\) on top of \(\triangle ABC\) so that \(D\) lies on \(AB\), \(F\) lies on the extension of \(BC\), \(DB=2\) cm, and sides \(AC\) and \(DF\) meet at \(E\). If \(AE=3\) cm and \(EC=2\) cm, determine the length of \(CF\).
Since \(AB=3\) and \(DB=2\), it follows that \(AD=1\) cm. Draw a perpendicular from \(E\) to \(BF\).
Let \(P\) be the point where the perpendicular intersects \(BF\).
Let \(CF = a\), \(PC = b\), and \(EP = h\).
We will now proceed with three solutions. The first two solutions depend on this setup. The first uses similar triangles, the second uses trigonometry, and the third uses coordinate geometry.
Solution 1
Since \(EP\) is perpendicular to \(BF\), we know \(\angle EPF = 90\degree\). Also, \(\angle ECP = \angle ACB\) (same angle). Therefore, \(\triangle ABC \sim \triangle EPC\) (by angle-angle triangle similarity).
From the similarity, \(\dfrac{AC}{BC} = \dfrac{EC}{PC}\), so \(\dfrac{5}{4} = \dfrac{2}{b}\) or \(b = \dfrac{8}{5}\). Also, \(\dfrac{AC}{AB} = \dfrac{EC}{EP}\), so \(\dfrac{5}{3} = \dfrac{2}{h}\) or \(h = \dfrac{6}{5}\).
Now let’s calculate \(PF\). We know \(\angle EPF = \angle DBF = 90\degree\) and \(\angle EFP = \angle DFB\) (same angle).
Therefore, \(\triangle DBF \sim \triangle EPF\) (by angle-angle triangle similarity). This tells us \(\dfrac{DB}{BF} = \dfrac{EP}{PF}\).
Since \(BF = BC + CF = 4+a\) and \(PF = PC + CF = \dfrac{8}{5} + a\), we have \[\begin{aligned} \frac{DB}{BF} &= \frac{EP}{PF}\\[1mm] \frac{2}{4 + a} &= \frac{\frac{6}{5}}{\frac{8}{5}+a}\\[1mm] \frac{16}{5}+2a &= \frac{24}{5}+\frac{6}{5}a\\[1mm] 2a - \frac{6}{5}a&= \frac{24}{5}-\frac{16}{5}\\[1mm] \frac{4}{5}a&= \frac{8}{5}\\ a &=2 \end{aligned}\] Therefore, \(CF = 2\) cm.
Solution 2
In \(\triangle EPC, ~ \sin({\angle ECP}) = \frac{h}{2}\). In \(\triangle ABC, ~ \sin({\angle ACB}) = \frac{3}{5}\).
Since \(\angle ECP = \angle ACB\) (same angle), \[\begin{aligned} \sin({\angle ECP})&= \sin({\angle ACB})\\ \frac{h}{2}& = \frac{3}{5}\\ h &= \frac{6}{5}\\ \end{aligned}\]
Since \(\triangle EPC\) is a right-angled triangle, \[\begin{aligned} EP^2 + PC^2 &= EC^2\\ h^2 + b^2 &= 2^2\\ \left(\frac{6}{5}\right)^2 + b^2 &= 4\\ b^2 &= 4 - \frac{36}{25}\\[1mm] b^2 &= \frac{64}{25}\\[1mm] b &= \frac{8}{5}, \quad \text{since} \quad b>0 \end{aligned}\]
In \(\triangle EPF, ~ \tan({\angle EFP}) = \dfrac{EP}{PF} = \dfrac{h}{a+b} = \dfrac{\frac{6}{5}}{a + \frac{8}{5}}\).
In \(\triangle DBF, ~ \tan({\angle DFB})= \dfrac{DB}{BF} = \dfrac{2}{4 + a}\).
Since \(\angle EFP= \angle DFB\) (same angle), \[\begin{aligned} \tan({\angle EFP})&= \tan({\angle DFB})\\ \frac{\frac{6}{5}}{a + \frac{8}{5}} &= \frac{2}{4 + a}\\[2mm] \frac{24}{5} + \frac{6}{5}a & = 2a + \frac{16}{5}\\[2mm] 2a - \frac{6}{5}a &= \frac{24}{5} - \frac{16}{5}\\[2mm] \frac{4}{5}a &= \frac{8}{5}\\[2mm] a & = 2 \end{aligned}\] Therefore, \(CF = 2\) cm.
Solution 3
We will use coordinate geometry in this solution, and place \(B\) at the origin. Using the given information, \(D\) is at \((0,2)\), \(A\) is at \((0,3)\), \(C\) is at \((4,0)\), and \(F\) is on the positive \(x\)-axis at \((f,0)\) with \(f>4\). Consider the circle through \(E\) with centre \(C(4,0)\). Since \(CE=2\), the radius of this circle is \(2\). Thus, the equation of this circle is \((x-4)^2+y^2=4\).
The line passing through \(A(0,3)\) and \(C(4,0)\) has \(y\)-intercept \(3\) and slope \(-\frac{3}{4}\), and so has equation \(y=-\frac{3}{4}x+3\). Since \(E\) lies on the line with equation \(y=-\frac{3}{4}x+3\) and the circle with equation \((x-4)^2+y^2=4\), to find the coordinates of \(E\), we substitute \(y=-\frac{3}{4}x+3\) for \(y\) in \((x-4)^2+y^2=4\). Note that \(E\) is in the first quadrant so \(x>0\) and \(y>0\).
Doing so, we get \[(x-4)^2+\left(-\frac{3}{4}x+3\right)^2=4\] Expanding the left side, we get \[x^2-8x+16+\frac{9}{16}x^2-\frac{9}{2}x+9=4\] Multiplying by \(16\), we get \[16x^2-128x+256+9x^2-72x+144=64\] Simplifying, we get \[25x^2-200x+336=0\] Factoring, we then get \[(5x-12)(5x-28)=0\] It follows that \(x=\frac{12}{5}\) or \(x=\frac{28}{5}\). Substituting \(x=\frac{12}{5}\) in \(y=-\frac{3}{4}x+3\), we obtain \(y=\frac{6}{5}\).
Substituting \(x=\frac{28}{5}\) in \(y=-\frac{3}{4}x+3\), we obtain \(y=-\frac{6}{5}\). But \(E\) is in the first quadrant so \(y>0\), and this second possibility is inadmissible. It follows that \(E\) has coordinates \(\left(\frac{12}{5},\frac{6}{5}\right)\).
We can now find the equation of the line containing \(D(0,2)\), \(E\left(\frac{12}{5},\frac{6}{5}\right)\), and \(F(f,0)\). This line has \(y\)-intercept \(2\), slope equal to \(\dfrac{\frac{6}{5}-2}{\frac{12}{5}-0} = \dfrac{-\frac{4}{5}}{\frac{12}{5}} = -\frac{1}{3}\), and thus has equation \(y=-\frac{1}{3}x+2\).
The point \(F(f, 0)\) lies on this line, so \(0 = -\frac{1}{3}(f) + 2\), which leads to \(f=6\). Thus, the point \(F\) has coordinates \((6,0)\). Since \(C\) is at \((4,0)\) and \(F\) is at \((6,0)\), \(CF=2\). It turns out that \(F\) also lies on the circle through \(E\).
Therefore, \(CF = 2\) cm.