Given rectangle $ABCD$ with $K$ the midpoint $AD$ and $AD/AB=\sqrt{2}$, find the angle between $BK$ and diagonal $AC$.

Question

Could someone help me with this little problem of geometry of quadrilaterals?

In rectangle $ABCD$, let $K$ be the midpoint of side $AD$. If we know that $AD/AB =\sqrt{2}$, find the angle between $BK$ and diagonal $AC$.

It is assumed that the problem can be solved using the Pythagorean theorem.

I have tried a little of everything but I can't find how to argue the answer well, I hope you can help me.

Answer 1

The angle between $BK$ and $AC$ is the difference in the angles each makes with the $x$ axis

Angle of $BK$ is $\phi_1 = \tan^{-1} \left( \dfrac{1}{\sqrt{2}/2} \right) = \tan^{-1} \left(\sqrt{2}\right)$

Angle of $AC$ is $\phi_2 = \pi - \tan^{-1} \left(\dfrac{1}{\sqrt{2}} \right) $

Hence the angle between them is

$ \phi =\phi_2 - \phi_1= \pi - \tan^{-1}\left(\dfrac{1}{\sqrt{2}}\right) - \tan^{-1}\left(\sqrt{2}\right) $

Taking the tangent of the angle $\phi$ , we get

$ \tan(\phi) = - \dfrac{ \dfrac{1}{\sqrt{2}} + \sqrt{2} }{ 1 - \dfrac{\sqrt{2}}{\sqrt{2}} } = \text{undefined} $

Since the tangent of the angle $\phi$ is undefined, then it follows that $\phi = 90^\circ$.

Answer 2

You don't need explicit trig. Since $AK=\frac {AD} 2$ we have:

$$\frac {AB}{AK} = 2 \frac {AB}{AD} =\sqrt 2$$

And you can proceed to show that the triangles $ABK$ and $ACD$ are similar (they are both right-angled). From which tracing equal angles gives you your answer.

Trigonometric ratios are (Amongst other things) a convenient way of encoding similarity. Sometimes pure geometry is simpler.