$ABCD$ is a square. $E$ is the midpoint of $CB$, $AF$ is drawn perpendicular to $DE$. If the side of the square is $2016$ cm , find $BF$.

Question

$ABCD$ is a square. $E$ is the midpoint of $CB$, $AF$ is drawn perpendicular to $DE$. If the side of the square is $2016$ cm , find $BF$.

What I Tried: Here is a picture,

I used a really peculiar way to solve this problem (I would say so). I found a lot of right-angled triangles in it and immediately used Pythagoras Theorem to find the length of $AE$ and $DE$ and found them to be $1008\sqrt{5}$ each .

Now I assumed $DF$ to be $x$, then $FE$ comes $(1008\sqrt{5} - x)$ .
From $AD$ and $DF$ again by Pythagoras Theorem I get $AF = \sqrt{2016^2 - x^2}$ .
Now comes the main part. From $AF$ and $EF$ along with $AE$ in right-angled $\Delta AFE$, I get :- $$(2016^2 - x^2) + (1008\sqrt{5} - x)^2 = (1008\sqrt{5})^2$$ $$\rightarrow 2016^2 - x^2 + (1008\sqrt{5})^2 - 2016\sqrt{5}x + x^2 = (1008\sqrt{5})^2 $$ $$\rightarrow 2016^2 = 2016\sqrt{5}x $$ $$\rightarrow x = \frac{2016}{\sqrt{5}}$$ From here I get $FE = \frac{3024}{\sqrt{5}}$ .

Now I used Ptolemy's Theorem on $\square AFEB$, noting that it is cyclic. $$AE * BF = (AB * EF) + (AF * BE) $$ $$ 1008\sqrt{5} * BF = (1008\sqrt{5} * 2016) + (\sqrt{2016^2 - \frac{2016^2}{\sqrt{5}}} - 1008^2)$$

Everything except $BF$ is known, so I am getting $BF$ as :- $$\frac{1270709}{630} - \frac{1008}{\sqrt{5}}$$ But to my surprise, the correct answer to my problem is simply $2016$ .

So my question is, was there any calculation errors? Or the method I used had a flaw in some way and so was not correct?

Can anyone help?

Alternate solutions are welcome too, but if someone can point out the flaw in my solution, then it will be better.

Answer 1

The error in your solution is in the Ptolemy Theorem step.

First, $AB\times EF$ is not $1008\sqrt{5} \times 2016$ because $EF$ is not $1008\sqrt{5}$, in fact as you calculated it is $\frac{3024}{\sqrt{5}}$.

Second, $AF\times BE$ is not $\sqrt{2016^2 - \frac{2016^2}{\sqrt{5}}} - 1008^2$, it should be $\sqrt{2016^2 - \frac{2016^2}{\sqrt{5}}} \times 1008$.

Alternatively, there is a pretty easy way: First since $\angle DAF=\angle CDE=\angle EAB$ we know $\angle DAE=\angle FAB$. Second since $A,F,E,B$ are co-cyclic we know $\angle FBA=\angle FEA$. Therefore triangles $\triangle DEA$ and $\triangle FBA$ are similar so $FB=AB$.

Answer 2

On Cartesian plane, take $A(0,0)$, $B(2016,0)$, $D(0,2016)$ etc.

$E=(2016,1008)$

Clearly equation of $ED$ (with slope $-1/2$) : $y=-x/2+2016$

Equation of $AF$ (slope $2$) : $y=2x$.

Find $AF\cap ED = F$. Now compute $BF$.

I solved for general $a$ (side of square) and got $$\boxed{BF=a}$$

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For your solution I suggest solving for variable side $2a$, to avoid big equations and numbers.

Apply Ptolemy to $ABEF$ ($AB=2a$, $BE=a$, $AE=a\sqrt{5}$, $AF=a\sqrt{5}$, $EF=\ldots$ No room for mistakes here.

Answer 3

You can find that $AF=2016\cdot \frac{2}{\sqrt{5}}$ since $\frac{1}{2}\cdot DE\cdot AF=\frac{1}{2}(2016)(2016)$. Now $\cos \theta=\frac{2}{\sqrt{5}}, \sin \theta =\frac{1}{\sqrt{5}}$. Thus, $GF=\frac{2}{5}\cdot 2016, AG=\frac{4}{5}\cdot 2016$. $FH=\frac{3}{5}\cdot 2016$, which yields $BF=2016$ right away.

Answer 4

Trigonometric Solution

$$\frac {BF}{\sin\angle FEB}=\frac{BE}{\sin\angle BEF}$$

$$\frac {BF}{\frac {2016}{1008\sqrt5}}=\frac{1008}{\frac{1008}{1008\sqrt5}}$$

$$BF=2016$$

$$\sin\angle FEB=\sin(180-\angle CED)=\sin\angle CED=\frac {2016}{1008\sqrt5}$$

$\angle BEF=\angle BAE$ (ABEF is cyclic quadrilateral)

$$\sin\angle BEF=\sin\angle BAE =\frac {1008}{1008\sqrt5}$$