# Problem of the Week Problem E and Solution Overlapping Shapes 3

## Problem

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 $$CF$$.

## Solution

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.