$ABC$ is a triangle such that $BC = 74 cm$ and point $D$ is on $BC$ so $BD=14cm$.If $\angle ADB=60^\circ$ then what is the area of triangle $ABC$?

In the figure below, $ABC$ is a triangle such that $\angle BAC = 150^\circ$ , $BC = 74 cm$ and point $D$ is on $BC$ such that $BD=14cm$.If $\angle ADB=60^\circ$,then what is the area, in $cm^2$, of triangle $ABC$?

I have no idea how to solve this, I think the height can somehow be calculated, but I dont know how, help aswell as solutions would be appreciated

Taken from the 2019 IMC

3 Answers

Construct the isosceles triangle ADE. Then, $\angle$AEC = 180 - $\frac 12\angle$ADB = 150. Thus, ACE and ACB are similar triangles with

$$AC^2=BC(DC - DE)\tag{1}$$

Apply the cosine law to the triangle ADC,

$$AC^2=AD^2+DC^2-AD\cdot DC\cos120\tag{2}$$

Let AD = DE = $a$. Eliminate AC in (1) and (2), and substitute other segments with $a$ and the given lengths,

$$a^2+60^2+60a=74(60-a)$$

which yields $a=6$. The area of the triangle is, then,

$$\frac 12 BC\cdot a\sin60= 111\sqrt 3$$

A path using similar triangles

Consider the Figure below, where $AH$ is the altitude of $\triangle ABC$ relative to $BC$, and three points on $BC$ have been taken so that $EF \cong FD \cong DG \cong AD$. Call $x$ the lenght of these segments.

1. Write $\overline{GC}$ as a function of $x$.
2. Observe that $\triangle AFG$ is right-angled and use this information to write $\overline{AG}$ as a function of $x$.
3. Use AAA criterion to demonstrate that $\triangle AGC \sim\triangle ABC$, and show therefore that $$\overline{AC}^2 = 74 (60-x),$$ and $$\overline{AB} = \frac{x\sqrt{3}\sqrt{74}}{\sqrt{60-x}}.$$
4. Demonstrate that $\triangle ABE \sim\triangle ABC$, which implies, in particular, that \begin{equation}\overline{BE} = \frac{3x^2}{60-x}.\tag{1}\label{eq1}\end{equation}
5. Observe that we can also write \begin{equation}\overline{BE} = 14-2x\tag{2}\label{eq2}.\end{equation} Equate \eqref{eq1} and \eqref{eq2}, and solve with respect to $x$. The positive solution is $x=6$.
6. Since $\overline{AH} = \frac{x\sqrt 3}2$, the required area is $$\mathcal A = \frac{74\cdot 6 \sqrt 3}{4}=111 \sqrt 3.$$

Let $AA'=:h$ be the height in $\Delta :=\Delta ABC$, where $A'\in BC$ is the foot of this height on the opposite side. Let $y,z$ be the (measures of) the angles in $B,C$ in $\Delta$. Then $y+z=30^\circ$.

Let $Y,Z$ be $Y=\cot y=BA':h$, $Z=\cot z=CA':h$. We obtain the following three equations in the unknowns $h,Y,Z$: $$ \left\{ \begin{aligned} \frac {YZ-1}{Y+Z} &= \sqrt 3\ ,\\ hY + \frac h{\sqrt 3} &= 14\ ,\\ hZ - \frac h{\sqrt 3} &= 60\ , \end{aligned} \right. \qquad\text{ from }\qquad \left\{ \begin{aligned} \frac {YZ-1}{Y+Z} &= \cot(y+z)=\cot 30^\circ\ ,\\ BA'+A'D &= BD=14\ ,\\ CA'-DA' &= CD=60\ . \end{aligned} \right. $$ The last two equations are linear in $Y,Z$, we express them linearly in terms of $h$, plug in into the first equation, so we get a quadratic equation in $h$ as stand alone variable.

Solving the system we (reject $h=-70\sqrt 3$, and) get the solution $h=3\sqrt 3$ (together with $Y=11\sqrt3 /9$, $Z=7\sqrt 3$), which leads to some explicit number which is the area of the triangle.