# Length of a side of a triangle with angle relations

In the image, $$AB=CD=5$$, $$BC=2$$. Then $$AD=?$$

My try

I extended $$BC$$ until intersecting $$AD$$, and i extended $$CD$$ until intersecting $$AB$$. After that i tried angle chasing inside the new triangles but i got nothing. Any hints?

This problem is meant to be resolved without trigonometry.

This looks like a trick question: the construction with all declared properties does not exist.

Let $$\angle BAC=\angle ADC=\alpha$$, $$\angle CBA=\beta$$, $$E=BC\cap AD$$ and let's ignore for the moment that we have a condition $$\angle CBA=2\angle BAC$$.

Then (as it was already noted in other answers) we must have

\begin{align} \angle ECA&=\angle EAC=\alpha+\beta ,\\ |ED|&=|AC|=|AE| ,\\ |CE|&=|BC|=2 . \end{align}

Let $$F$$ be the median (and the altitude) of the isosceles $$\triangle AEC$$. Then we must recognize $$\triangle AFB$$ as the famous $$3-4-5$$ right-angled triangle, with $$|AF|=4$$. Now we can easily find $$|AE|$$ and $$|AD|$$:

\begin{align} \triangle EFA:\quad |AE|&=\sqrt{|AF|^2+|FE|^2}=\sqrt{17} ,\\ \end{align}

and we are tricked to state that we have the answer: \begin{align} |AD|&=2|AE|= 2\sqrt{17} . \end{align}

Ant it would be indeed the correct answer, in case when we do not have any relations between the angles $$\alpha$$ and $$\beta$$ .

Otherwise we must check that the other declared conditions hold, that is, \begin{align} \angle BAC=\alpha&=x ,\\ \angle CBA=\beta&=2x , \end{align}

in other words

\begin{align} \angle CBA&=2\angle BAC . \end{align}

Let's check if the following is true:

\begin{align} \triangle AFC:\quad\alpha+\beta& =x+2x=3x=\arctan4 ,\\ \triangle AFB:\phantom{\quad\alpha+}\beta &=2x = \arctan\tfrac43 . \end{align}

So, the following must be true: \begin{align} \tfrac13\arctan4&= \tfrac12\arctan\tfrac43 , \end{align} but this is false, since \begin{align} \tfrac13\arctan4 &\approx 0.4419392213 ,\\ \tfrac12\arctan\tfrac43 &\approx 0.4636476090 , \end{align}

thus the original question does not have a valid solution.

Edit Another illustration that given geometric construction is invalid.

Triangle $$ABC$$ with given constraints on the angles and side lengths uniquely defines $$x$$:

\begin{align} \triangle ABC:\quad \frac{|BC|}{\sin x}&= \frac{|AB|}{\sin 3x} ,\\ \frac{2}{\sin x}&= \frac{5}{\sin x \,(3-4\sin^2x)} ,\\ \sin^2x&=\tfrac18 ,\\ \sin x&=\tfrac{\sqrt2}4 ,\\ x&=\arcsin\tfrac{\sqrt2}4\approx 20.7^\circ , \end{align}

Given $$x$$, we can construct $$\triangle ABC$$ and the point $$D$$, but the ray $$DX:\angle XDC=x$$ will miss the point $$A$$, and the real picture should in fact look like this:

The idea of this answer is the same as the one in Lee's answer. One can get $$AD$$ without using trigonometry.

Let $$E$$ be the intersection point of $$AD$$ with $$BC$$.

Then, since $$\angle{ACE}=3x$$, we get $$\angle{DCE}=5x-3x=2x$$.

It follows that $$\triangle{CBA}\equiv\triangle{ECD}$$. So, $$CE=2$$ and $$ED=CA$$.

Since $$\angle{CEA}=x+2x=3x$$, we see that $$\triangle{ACE}$$ is an isosceles triangle. It follows that $$AE=AC$$.

Let $$AC=p$$. Then, using $$CA^2+CD^2=2(CE^2+AE^2)\tag1$$ (see here) one gets $$p^2+5^2=2(2^2+p^2)\implies p=\sqrt{17}$$ from which $$AD=2p=2\sqrt{17}$$ follows.

The wiki page uses trigonometry for proving $$(1)$$.

One can prove $$(1)$$ without using trigonometry.

Let $$\vec{EC}=\vec c,\vec{EA}=\vec a$$. Then, $$\vec{ED}=-\vec a$$ and \begin{align}CA^2+CD^2&=|\vec{a}-\vec{c}|^2+|-\vec{a}-\vec{c}|^2 \\\\&=2|\vec a|^2+2|\vec c|^2 \\\\&=2EA^2+2EC^2\qquad\square\end{align}

Lets say $$BC$$ intersects $$AD$$ at point $$E$$, then $$ABC$$ is similar to $$DCE$$, those $$AC=ED$$ and $$CE=2$$. Also notice that $$ACE$$ is isosceles, thus $$AC=ED=AE$$, i.e. $$AD=2AC$$.

I don't know, but maybe it helps to solve your problem without trigonometry