How to construct a triangle given two sides and their bisector?

Question

Suppose I have two triangle sides $AB$ and $AC$, and the length of the angle bisector of $A$. How can I construct (straightedge and compass) the triangle? (This question is from one of the earlier Moscow Math Olympiads.)

Answer 1

Let the length of the angle bisector of $A$ $|AD|=d$ and the side lengths $|BC|=a,|AC|=b,|AB|=c$, $|BD|=m$.

Then according to Stewart's Theorem for $\triangle ABC$,

\begin{align} c^2(a-m)+b^2m -a(d^2+(a-m)m) &=0 \tag{1}\label{1} . \end{align}

By the law of sines,

\begin{align} \triangle ABD:\quad \frac{m}{\sin\tfrac\alpha2} &= \frac{c}{\sin\delta} ,\\ \triangle ADC:\quad \frac{a-m}{\sin\tfrac\alpha2} &= \frac{b}{\sin\delta} ,\\ \text{hence, }\quad \frac{m}{a-m}&=\frac{c}{b} ,\\ a&=\frac{m(c+b)}{c} , \end{align}

and \eqref{1} becomes

\begin{align} \frac{m(c+b)(c^2b-m^2b-d^2c)}{c^2} &=0 ,\\ \end{align}

\begin{align} m^2&= \frac{c(bc-d^2)}{b} . \end{align}

Given $m$ we can found that

\begin{align} a^2&=\frac{(bc-d^2)(b+c)^2}{bc} , \end{align}

hence, the length of the missing side $a$ can be constructed from known values $b,c,d$, for example, using Intersecting_chords_theorem

Answer 2

By Stewart's theorem the length $\ell$ of the angle bisector through $A$ fulfills $$ \ell^2 =\frac{bc}{(b+c)^2}\left((b+c)^2-a^2\right)$$ hence $$a\sqrt{bc}= (b+c)\sqrt{bc-\ell^2}$$ and it is enough to construct three segments with lengths $\sqrt{bc},(b+c),\sqrt{bc-\ell^2}$ to find $a$ (then $ABC$) through the intersecting chords theorem:

Answer 3

We can find the length of side BC as follows. Let $AD = m$ be the angle bisector such that $D$ is on $BC$ and $\angle BAB = \angle CAD = \theta$ . Since $AD$ is angle bisector of $\angle A $ therefore $${AB \over AC} = {BD \over DC} \implies \frac cb = {BD\over DC} $$ Let $BD = kc$ and $DC = kb$. Now using cosine rule in $\Delta CAD$ and $\Delta BAD $ we get $$(kb)^2 = b^2 + m^2 - 2bm \cos \theta $$ and $$ (kc)^2 = c^2 + m^2 - 2cm \cos \theta $$ Now equating $\cos\theta$ in both equation, we get $$ {k^2b^2-b^2-m^2 \over -2bm} = {k^2c^2-c^2-m^2 \over -2cm}$$ Simplyfying the above equation we will get $$ k = \sqrt{{bc - m^2 \over bc}} $$ So $$BC = \sqrt{{bc-m^2 \over bc}}(c+b)$$ Now we have all sides of $\Delta ABC$.

Answer 4

A simple and unique method to solving this is by the proof of the angle bisector theorem by use of parallel lines. Here’s how it goes.

Lets us denote side AB by c and AC by b. and the bisector of angle A as t.

• Construct a triangle PAC having sides PA=AC=b and PC = t/c * (b+c) by SSS basic construction. [There are many articles online on how to do SSS construction.]
• Extend PA to B such that AB=c and join BC.

Here’s a little insight on why it is true. Consider the auxiliary figure for the above construction:

• Let the line t intersect BC at D
• Draw a line parallel to t from AC. Let AB intersect this line at P.
• By AAA similarity, triangle BAD is similar to triangle BPC.
• By ratio of equal sides and isosceles triangle theorems: c/t = (b+c)/PC Transposing we obtain PC = t/c * (b+c)
• And the rest of the construction is as given above. Hope this helped.