If $\overline{AD}=\overline{DC}$, $\angle ABC=30^\circ$, $\angle ADC=60^\circ$, $\overline{AB}=55$ and $\overline{CB}=48$, what is $\overline{BD}$?

Question

I was posed the following geometry question by a friend of mine. She mentioned that there was a very pretty solution that avoided trig. I’m pretty bad at geometry, however, so I could only find a solution using trig.

Let $\overline{AB}=55$, $\overline{BC}=48$, $\overline{AD}=\overline{DC}$, $\angle ABC = 30^\circ$, and $\angle ADC = 60^\circ$. Find $BD$.

First, since $\overline{AD}=\overline{DC}$, we can quickly establish that $\triangle ADC$ is Isosceles. Since all isosceles triangles with a $60^\circ$ angle are regular, we get that $\triangle ADC$ is equilateral and thus $\overline{AC}=\overline{DC}=\overline{AD}$.

Next, I discovered a fairly interesting fact about the construction that led nowhere. If we construct $\triangle ABC$’s circumcircle (which I will call $\Gamma_1$, and another circle centered at $D$ that intersects $A$ and $C$ (which I will call $\Gamma_2$) , we see by the inscribed angle theorem that arc $AC$ has the same degree measure on both $\Gamma_1$ and $\Gamma_2$, and thus the radii of $\Gamma_1$ and $\Gamma_2$ are equal.

However, I didn’t find anything useful about these circles, so I gave up on them

After being stuck for a while, I admitted defeat and tried using trig:

By the law of cosines, $\overline{AC}=\sqrt{55^2+48^2-2\cdot 55 \cdot 48 \cdot \cos(30^\circ)}=\sqrt{5329-2640\sqrt{3}}$.

Next, we let $\angle BAC=\theta$. Then by extended law of sines, we see that $\tfrac{48}{\sin(\theta)}=d$ where $d$ is the circumdiameter of $\triangle ABC$, or the diameter of $\Gamma_1$. As shown previously, this is also the diameter of $\Gamma_2$, which is $2\overline{AD}=2\overline{AC}=2\sqrt{5329-2640\sqrt{3}}$. This tells us that $\sin (\theta)=\tfrac{24}{\sqrt{5329-2640\sqrt{3}}}$. We can also use this to compute $\cos (\theta)$ So we have

$$\sin (\theta) = \frac{24}{\sqrt{5329-2640\sqrt{3}}}$$ $$\cos(\theta) = \sqrt{\frac{4753-2640\sqrt{3}}{5329-2640\sqrt{3}}}$$

Now, consider $\triangle ADB$. We know $\overline{AD}=\sqrt{5329-2640\sqrt{3}}$, $\overline{AB}=55$, and $\angle DAB= \theta + 60^\circ$, so we can use law of cosines to get $\overline{DB}$.

$$\overline{DB}^2= \overline {AD}^2 + \overline{AB}^2 - 2\overline {AD} \cdot \overline {AB} \cos(\theta + 60 ^ \circ)$$ $$= 5329 - 2640\sqrt{3} + 3025 -110\sqrt{5329-2640\sqrt{3}}(\cos(\theta)\cos(60^\circ)-\sin(\theta)\sin(60^\circ))$$ $$=8354-2640\sqrt{3}-55\sqrt{5329-2640\sqrt{3}}\Bigr(\sqrt{\frac{4753-2640\sqrt{3}}{5329-2640\sqrt{3}}} -\sqrt{3} \frac{24}{\sqrt{5329-2640\sqrt{3}}} \Bigr)$$ $$=8354-2640\sqrt{3}-55(\sqrt{4753-2640\sqrt{3}}-24\sqrt{3})$$ $$=8354-2640\sqrt{3}-55((55-24\sqrt{3})-24\sqrt{3})$$ $$=8354-55^2-2640\sqrt{3}+55\cdot 48\cdot \sqrt{3}$$ $$=5329$$

So $\overline{DB}=\sqrt{5329}$. Interestingly, $\sqrt{5329}=\sqrt{55^2+48^2}$, which is a very simple form, and indicates how this problem might be generalized.

However, my friend said that this problem has a solution without trig, and this definitely uses a lot of trig. So my question is: how could I solve this problem without using trig?

Answer 1

Hint :

Construct equilateral triangle $BCE$ where $E$ lies on opposite side of $A$.

Find a suitable pair of congruent triangles.

$$\triangle BCD \cong \triangle ECA \quad (SAS) $$

Thus $BD=AE$ where $AE$ is easy to compute.