Question

In the rectangle ABCD shown below, AB=4 and BC=2. Let E be the midpoint of the side BC, and let AE cut BD at F. Prove that ∠BCF=45°.

Answer 1

Given the figure:

We know that AB = 4, so we draw the height from F to AB, and we call it h:

From the figure, we know that:

`\begin{gathered} \tan \theta=(BE)/(AB)=(1)/(4)=(h)/(AM)\Rightarrow AM=4h \\ \tan \alpha=(AD)/(AB)=(2)/(4)=(h)/(MB)\Rightarrow MB=2h \end{gathered}`

From this, we can say that:

`\begin{gathered} 4h+2h=4 \\ h=(2)/(3) \\ \Rightarrow FN=(4)/(3) \\ \Rightarrow EN=1-h=1-(2)/(3)=(1)/(3) \\ \Rightarrow CN=1+EN=(4)/(3) \end{gathered}`

Then:

`\tan x=(FN)/(CN)=(4/3)/(4/3)=1`

We conclude that x must be 45° because tan(45°) = 1.