Find angle in triangle $ABC$ with cevian line $AD$, such that $AB=CD$.

Question

As you can see in the picture, there is a triangle $ABC$ with $∠C=30°$ and $∠B=40°$. Now we assuming that $AB=CD$, try to find the exact value of $∠CAD$.

My attempt: Denote $∠CAD$ by $x$, we know that $$\frac{\sin C}{AD}=\frac{\sin{x}}{CD},\quad\frac{\sin B}{AD}=\frac{\sin{(x+C)}}{AB}$$ Then we have ($∠C=30°=\frac{\pi}{6},∠B=40°=\frac{2\pi}{9}$) $$\frac{\sin{\frac{\pi}{6}}}{\sin{x}}=\frac{\sin{\frac{2\pi}{9}}}{\sin{(x+\frac{\pi}{6})}}=\frac{AD}{AB}$$

Everything looks ok so far, but I have trouble solving the equation. What's more, Wolfram tells me that the answer is $x=\frac{5\pi}{18}$.

This exercise is in my sister's assignment, so I think this exercise should have a high-school (or high-school olympic) level answer.

More: The exercise appears in geometry part, so a pure geometric method will be better.

Answer 1

By your work: $$\sin(x+30^{\circ})=2\sin40^{\circ}\sin{x}$$ or $$\sin{x}\cos30^{\circ}+\cos{x}\sin30^{\circ}=2\sin40^{\circ}\sin{x}$$ or $$\tan{x}=\frac{1}{2(2\sin40^{\circ}-\cos30^{\circ})}.$$ But, $$\frac{1}{2(2\sin40^{\circ}-\cos30^{\circ})}=\frac{1}{2(\sin40^{\circ}+\sin40^{\circ}-\sin60^{\circ})}=$$ $$=\frac{1}{2(\sin40^{\circ}-2\sin10^{\circ}\cos50^{\circ})}=\frac{1}{2\cos50^{\circ}(1-2\sin10^{\circ})}=$$ $$=\frac{\tan50^{\circ}}{2\sin50^{\circ}(1-2\sin10^{\circ})}=\frac{\tan50^{\circ}}{2(\sin50^{\circ}-\cos40^{\circ}+\cos60^{\circ})}=\tan50^{\circ},$$ which says $x=50^{\circ}.$

Answer 2

You can do it also syntheticly.

Draw an equilateral triangle $ABF$ where $F$ is on different side of a line $BC$ then $A$. Let $AE$ cuts $BC$ at $D'$ and prove $D=D'$ i.e. $CD' = AB$.

Note that since $\angle BCA = {1\over 2}\angle BFA $ the point $C$ lies on circle centered at $F$ and $r=FA = FB$. So $FC = AB$. Now easy angle chase we see that $\angle CD'F= \angle = CFD' = 80^{\circ}$ so $CF = CD'$ and you are done.

Answer 3

After asking a middle school math teacher, I got the answer as following.

Make an equilateral triangle $EBC$, and then connect $EA$ and make $DF/\!\!/EC$ which intersects $BE$ at point $F$. Now we have the picture above.

Notice that $∠BCA=∠ECA=30°$, we have $AB=AE$. Then from $∠ABC=40°$, we are able to get that $$∠EAB=180°-2∠AEB=180°-2(60°-∠ABC)=140°$$ From $DF/\!\!/EC$, we could obtain that $EF=CD=AB=AE$, which implies that $$∠EAF=∠EFA=\frac{180°-∠AEB}{2}=80°$$ Then $A,D,B,F$ are in a circle, which implies that $∠DAB=∠DFB=60°$ and the answer is clear.

Answer 4

An algebraic proof.

Consider the triangle $ACD$ and label $\varphi=\angle\, CAD$. Both $AC$ and $AD$ can be related to $AB$, i.e., $CD$, by the law of sines: $\frac{CD}{\sin \angle CAD}=\frac{AD}{\sin \angle ACD}=2AD$, thus $AD=\frac{AB}{2\sin \varphi}$. Also, $\frac{AC}{\sin \angle ABC}=\frac{AB}{\sin \angle ACB}=2AB$, thus $AC=2\sin 40^\circ AB$. By the law of cosines,

$$CD^{2}=AC^{2}+AD^{2}-2AC\cdot AD\cos\varphi \\ \Rightarrow \left(4\sin^{2}40^\circ+\frac{\csc^{2}\varphi}{4}-1-{2\cot\varphi\sin40^\circ}\right)AB^{2}=0 $$

Since $AB>0$, this equation has a solution iff $\varphi$ is a root of the parenthetical term. Let $\cos\varphi=x$, $\sin40^\circ=\lambda$. Thus, equating the parenthetical term to zero gives the quadratic in $x^2$,

$$16\left(16\lambda^{4}-4\lambda^{2}+1\right)x^{4}+8\left(-64\lambda^{4}+20\lambda^{2}-3\right)x^{2}+\left(256\lambda^{4}-96\lambda^{2}+9\right)=0$$

which can be solved for

$$\begin{align} x&=\pm\frac{1}{4}\sqrt{\frac{256\lambda ^{4}-80\lambda ^{2}+12\pm\ 8\sqrt{3}\lambda }{\left(16\lambda ^{4}-4\lambda ^{2}+1\right)}} \\ x&\in\left\{\pm0.6428,\,\pm0.9740\right\} \\ \varphi&\in\left\{13.12^\circ,\,50.00^\circ,\,130.00^\circ,\,166.91^\circ\right\} \end{align} $$

It suffices to see that $\varphi=50.00^\circ$ is the correct solution, since the angle is acute and $\varphi=13.12^\circ$ would make $\angle\,DAB=96.88^\circ$, which is absurd.