Relationship between two triangle side lengths where one other side is shared This question was asked on an Australian year 10 (15 to 16 year olds) practice exam.
Diagram of two triangles with sides a and b indicated:

"Determine the relationship between the values of $a$ and $b$ by writing $a$ in terms of $b$". The solution given was simply the following.
$a=\dfrac{b}{b \sqrt 3 - 1} \tag{1}\label{1}$
My attempt to solve this used the cosine rule on each of the two smaller triangles to get the side length opposite the $30°$ angle, then on the larger triangle for the side length opposite the $60°$ angle, giving the following relationship between a and b.
$\left(\sqrt{a^2 + 1 - a \sqrt 3} + \sqrt{b^2 + 1 - b \sqrt 3}\right)^2 = a^2 + b^2 - ab \tag{2}\label{2}$
However I was not able to simplify (\ref{2}) to get equation (\ref{1}).
My question is:

*

*How can equation (\ref{2}) be simplified to give equation (\ref{1}) using algebra that is accessible to a high school student?

*Is there another way, perhaps using other trigonometric identities, that does not use the form of (\ref{2})?

 A: 
The answer to your $2^\text{nd}$ question is “Yes”. But, to express $a$ in terms of $b$, you need to add two lines to your diagram and extend an existing segment. First, construct the perpendicular to the side $CA$ through the vertex $C$ to cut the extended side $AB$ at $P$. Then, draw a parallel line $CQ$ to $CP$ through the point $D$ to meet the extended side $AB$ at $Q$.
As shown in the diagram, the segment $CP$ makes $\triangle APC$ a right-angled-triangle, while segment $DQ$ makes $\triangle AQD$ an isosceles triangle.
Using the properties of a right-angled-triangle, we shall write, $\space AP=\dfrac{AC}{\cos\left(60^o\right)}=2b$.
Therefore, we have, $BP=AP-AB=2b-a$.
Since $AD$ is the $\hat{A}$-angle-bisector, we can apply angle bisector theorem to $\triangle ABC$ to obtain,
$$\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{a}{b}$$
Now, consider the triangle $BPC$, where $DQ$ is parallel to its side $CP$. Therefore, we have,
$$\dfrac{BQ}{QP}=\dfrac{DB}{DC}=\dfrac{a}{b},$$
with which we deduce that,
$$\dfrac{BQ}{BP}=\dfrac{BQ}{BQ+QP}=\dfrac{a}{a+b}\qquad\rightarrow\qquad BQ=\dfrac{a}{a+b}BP=\left(\dfrac{a}{a+b}\right)\left(2b-a\right).\tag{2}$$
According to the properties of an isosceles triangle, we know that, in $\triangle AQD$,
$$AQ= 2AD\cos\left(30^o\right)=\sqrt{3}.\tag{3}$$
Using (2), $AQ$ can be expressed also as,
$$AQ=AB+BQ=a+\left(\dfrac{a}{a+b}\right)\left(2b-a\right).\tag{4}$$
When we equate (3) and (4), we get the equation needed for determining $a$ in terms of $b$.
$$a+\left(\dfrac{a}{a+b}\right)\left(2b-a\right)=\sqrt{3}.$$
A: 
As suggested by aschepler,
sum of areas of the 2 smaller triangles = area of the larger triangle.
$$\frac{1}{2}(a)(1)\sin 30^o+\frac{1}{2}(1)(b)\sin 30^o=\frac{1}{2}(a)(b)\sin 60^o$$
$$\frac{1}{2}(a)(1)\left(\frac{1}{2} \right)+\frac{1}{2}(1)(b)\left(\frac{1}{2} \right)=\frac{1}{2}(a)(b)\left(\frac{\sqrt 3}{2} \right)$$
$$a+b=ab\sqrt 3$$
$$\therefore  a=\frac{b}{b \sqrt 3-1}$$
A: I would like to further add that this problem can be solved fairly simply using the method below:

1.) First, we label the triangle $\triangle ABC$ with a point $D$ that lies on $AC$ for our ease. We can also labor $\angle ACB=\alpha$. Now, we draw a line $DE$ from $D$ onto a point $E$ that lies on $AB$, such that $DE\parallel BC$.
2.) This gives us an isosceles triangle $\triangle BED$ where $\angle EDB=\angle EBD=30^\circ$. We also know that $BE=ED$ as well, furthermore, we can calculate the length of $BE$ and/or $ED$ as we know the length of $BD$, there are multiple ways of doing this (law of cosines, dropping a perpendicular, etc...), but it's fairly trivial so I'll just use the result and say that $BE=ED=\frac{1}{\sqrt3}$. This also implies that the remaining line segment $EA=b-\frac{1}{\sqrt3}$ or $\frac{b\sqrt3 -1}{\sqrt3}$.
3.) With this information now, we can finish off the problem. Notice that $\triangle AED$ is similar to $\triangle ABC$ via the AAA property. In that case, we can use the Thales theorem and say that:
$$\frac{1}{a\sqrt3}=\frac{b\sqrt3 -1}{b\sqrt3}$$
$$\frac{1}{a}=\frac{b\sqrt3 -1}{b}$$
Now we can just "flip" the expressions and get the desired result:
$$a=\frac{b}{b\sqrt3 -1}$$
To answer your "first" question, it may be possible to simplify that algebraic expression and find an equation for $b$ in terms of $a$, but as you stated too, this is not something that would typically be expected of a high-schooler. Even if you can simplify it, its simply not an efficient way to solve the problem.
