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.

enter image description here


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}

enter image description here

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: enter image description here


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


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