Find the missing angle formed by the intersection of a diagonal and an equal line inside the square.

Question

(Apologies for the vague title, I could not figure out how to describe it adequately within the word limit)

I came across this problem in an online forum a few months ago, with no answers under it. The diagram looks like this:

We have a square and another right angle at its lower left vertex, and the question is to find the missing angle formed by the intersection of the diagonal of this square, and another line segment of equivalent length. This geometry problem is quite interesting and unique compared to what I’ve posted before, there is not a lot of information given.

I will post my solution below as an answer. Please let me know if my answer is correct, and if there are any problems in my approach. Also, please post your own answers as well, preferably using different methods in order to ensure my answer is accurate.

Answer 1

Consider the diagram

Since $\angle PAD$ and $\angle QAB$ are both complementary to $\angle DAQ$, $$ \angle PAD=\angle QAB\tag1 $$ Thus, by SAS, we have $$ \triangle PAD\simeq\triangle QAB\tag2 $$ Note that $\angle PDA=\angle QBA=\frac\pi4$. Therefore, $$ \begin{align} \angle PDQ&=\frac\pi2\tag{3a}\\ \angle PDC&=\frac{3\pi}4\tag{3b} \end{align} $$ The Law of Sines says that $$ \begin{align} \sin(\angle DPC) &=\frac{DC}{PC}\,\sin(\angle PDC)\tag{4a}\\[3pt] &=\frac1{\sqrt2}\frac1{\sqrt2}\tag{4b}\\ &=\frac12\tag{4c} \end{align} $$ Since $\angle DPC=\frac\pi6$ and $\triangle RDP$ is a right triangle, we get that $\angle DRP=\frac\pi3$. Therefore, since vertical angles are equal $$ \angle BRC=\frac\pi3\tag5 $$

Answer 2

This answer I’m going to post will be a little different to my usual answers. This time, instead of writing an explanation, I shall post a Proof-Without-Words as following:

Please let me know if any clarification is needed. I’m happy to add some explanation to my answer iff necessary

Answer 3

One way I think about these problems with little or almost no information given apart from the figure, is trying to figure out why such a configuration would be unique.

Consider $\angle RAB =\theta$. Let us throw the figure onto a coordinate plane with A as origin and sides AB and AC as coordinate axes. We have chosen the side length as 1, since the problem asks for only an angle; any square with such a configuration would be similar with $ABCD$ and the angle would thus be conserved.

Then, the equation for line $CD$ is $x+y=1$ and for $AR$ is $y=x\tan\theta$. This means that the coordinates of $R$ are $\bigg(\dfrac{1}{1+\tan\theta}, \dfrac{\tan\theta}{1+\tan\theta}\bigg).$

This implies that the length of $AR$ is $\displaystyle \sqrt{\bigg(\dfrac{1}{1+\tan\theta}\bigg)^2+\bigg(\dfrac{\tan\theta}{1+\tan\theta}\bigg)^2} = \dfrac{\sec\theta}{1+\tan\theta}.$

Now,since $AT$ and $AR$ are perpendicular, the product of their slopes must equal $-1$, so the equation for $AT$ is $x = -y\tan\theta$. Now suppose $T = (-u\tan\theta, u)$. Then the length $$AT = \sqrt{u^2+u^2\tan^2\theta}=u\sec\theta$$ which should equal $AR=\dfrac{\sec\theta}{1+\tan\theta}$ so that $$u = \dfrac{1}{1+\tan\theta} \ \text{and}\ T = \bigg(-\dfrac{\tan\theta}{1+\tan\theta}, \dfrac{1}{1+\tan\theta}\bigg).$$

Now the final step is to see when the lengths $TD$ and $CB$ will be equal. Hopefully this will yield just one acute value of the angle $\theta$.
Equating the lengths, we get $$\sqrt{\bigg(\dfrac{1+2\tan\theta}{1+\tan\theta}\bigg)^2+\bigg(\dfrac{\tan\theta}{1+\tan\theta}\bigg)^2} = \sqrt2.$$solving which we get $\tan^2 \theta = \dfrac13$, from which the acute value is $\fbox{$\theta=30^\circ$}$.

Now, the slope of $DT$ is $\dfrac{\tan\theta}{1+2\tan\theta}=\dfrac{1}{2+\sqrt3}=2-\sqrt3$, so the angle $DT$ would have made with a horizontal line is $\arctan(2-\sqrt3) = 15^\circ.$ Thus the acute angle $DT$ makes with a vertical line is $90^\circ - 15^\circ = 75^\circ$.

Thus in $\triangle DXB$, we have $\angle BDX = 75^\circ$ and $\angle DBX=45^\circ$, which gives us $\bbox[7px,border:4px solid red]{\angle DXB = 60^\circ}.$

Answer 4

Comment: your construct is due to the value you know. In this cas we can also construct like this:

As shown in figure F, the intersection of BD and CE, is on pependicular bisector of EG and we have:

$\triangle AEF=\triangle AFG\Rightarrow \angle EFA=\angle AFB$

$\triangle BFC=\triangle BFA\Rightarrow \angle BFC=\angle AFB$

Therefore:

$\angle BFC=\angle AFB=\angle AFE=\frac{180}3=60^o$