Geometric construction of a triangle, provided an angle, an internal angular bisector of this angle, and the length of the side opposite to this angle

Question

How can I construct a triangle knowing the following information below?

A) An angle.

B) the length of the internal bisector for the given angle.

C) the length of the side opposite to the given angle.

Answer 1

This is a proof that, if such a triangle exists, then it can be constructed with a straightedge and a compass, and it is uniquely determined by the given three parameters. I may come back later for an actual construction. (Edit: A construction is given in a separate answer.)

Let $ABC$ be the triangle with given $\angle ACB=:2\gamma$, $AB=:c$, and $CD=:t$, where $D$ is the interior point of $AB$ such that $CD$ bisects $\angle ACB$. Write $a:=BC$ and $b:=CA$. Then, we have that $$a^2+b^2-2ab\,\cos(2\gamma)=c^2\tag{1}$$ and it is relatively well known that $$t^2+AD\cdot DB=ab\,.\tag{2}$$ Since $AD=\dfrac{bc}{a+b}$ and $BD=\dfrac{ac}{a+b}$, we get from (2) that $$t^2=\frac{ab}{(a+b)^2}\big((a+b)^2-c^2\big)=\frac{4a^2b^2}{(a+b)^2}\,\left(\frac{s(s-c)}{ab}\right)=\left(\frac{2ab}{a+b}\right)^2\,\cos^2(\gamma)\,,$$ where $s:=\dfrac{a+b+c}{2}$. Ergo, $$t=\frac{2ab}{a+b}\,\cos(\gamma)\,.\tag{3}$$ From (1), we see that $$(a+b)^2-4ab\,\cos^2(\gamma)=c^2\,,$$ whence, using (3), we get $$(a+b)^2-2t\,\cos(\gamma)\,(a+b)=c^2\,.$$ Thus, we have $$a+b=t\,\cos(\gamma)+\sqrt{t^2\,\cos^2(\gamma)+c^2}\,.\tag{4}$$ From (3) and (4), we obtain $$ab=\frac{t}{2\,\cos(\gamma)}\,\left(t\,\cos(\gamma)+\sqrt{t^2\,\cos^2(\gamma)+c^2}\right)\,.\tag{5}$$ Without loss of generality, we assume that $a\geq b$, and so (4) and (5) imply that $$a-b=\sqrt{c^2-2t^2\sin^2(\gamma)-\frac{2t\,\sin^2(\gamma)}{\cos(\gamma)}\,\sqrt{t^2\,\cos^2(\gamma)+c^2}}\,.\tag{6}$$ Since addition, subtraction, length multiplication, length division, length squaring, taking the square root of a length, halving a length, halving an angle, taking the cosine of an angle, and taking the sine of an angle are constructible processes, we conclude that $a$ and $b$ can be constructed, provided that $a-b$ as found in (6) is a real number. Thus, if $a-b$ is a real number, then the triangle $ABC$ can be constructed using a straightedge and a compass. Because $a$ and $b$ are uniquely determined by (4) and (6), the triangle $ABC$ is unique (up to congruence).

Answer 2

Because MathJax is being slow, I have to put this in a separate answer. Recall my notations from my previous answer. Now, I shall give a construction of the triangle $ABC$.

Step I. Construct the circumcircle $\Gamma$ of the triangle $ABC$. This is easy. Simply create an angle $AXY$ of size $2\gamma$ such that $AX<c$. Use $A$ as the center of a circle with radius $c$. This circle meets the ray $XY$ at $B$. Now, $\Gamma$ is just the circumcircle of the triangle $AXB$. Let $M$ be the midpoint of the arc $AB$ not containing $X$.

Step II. Create a line segment $PQ$ of size $t$. Let $R$ be a point such that $\angle PQR=\gamma$ and $\angle QRP=\dfrac{\pi}{2}$. Note that $QR=t\,\cos(\gamma)$. Extend $PR$ to a point $S$ such that $RS=c$. Then, $$QS=\sqrt{t^2\,\cos^2(\gamma)+c^2}\,.$$ Draw a circle centered at $R$ with radius $QS$. This circle meets the ray $QR$ at $T$. We have $$QT=QR+QS=t\,\cos(\gamma)+\sqrt{t^2\,\cos^2(\gamma)+c^2}=a+b\,.$$ Draw a perpendicular to $QT$ at $T$. Let $U$ be the point of intersection between this perpendicular line and the ray $QP$. The point $V$ denote the midpoint of $QU$. Observe that $$QV=\frac{QT}{2\,\cos(\gamma)}=\frac{a+b}{2\,\cos(\gamma)}\,.$$

Step III. From $t^2+AD\cdot DB=ab$, we get that $$CM=CD+DM=t+\frac{AD\cdot DB}{t}=\frac{ab}{t}=\frac{a+b}{2\,\cos(\gamma)}=QV\,.$$ Thus, use $M$ as the center of a circle with radius $QV$. This circle will intersect $\Gamma$ at two points (or one point in the degenerate case, or zero point when the parameters are impossible), and any of such point can be taken as our desired point $C$.

Answer 3

Draw isosceles $\triangle AHG$ with bisector $|AD|=d$

\begin{align} |CD|=|LD|&=u ,\\ |BD|=|KD|&=v ,\\ u+v&=a . \end{align}

The line $AD$ is tangent at the point $D$ to two ellipses, defined by with the constant sum $a$ and the foci $L,B$ and $K,C$.

We can also find \begin{align} g&=|HD|=|GD|=d\tan\tfrac\alpha2 ,\\ |FD|&=d\sin\tfrac\alpha2 ,\\ \cos\angle HDF&=\frac{|FD|}{g} ,\\ u&=\frac{|FD|}{\cos(\angle HDF-\theta)} ,\\ v&=\frac{|FD|}{\cos(\angle HDF+\theta)} . \end{align}

This provides the equation for $\theta$ \begin{align} \frac{|FD|}{\cos(\angle HDF-\theta)}+ \frac{|FD|}{\cos(\angle HDF+\theta)} &=a ,\\ \end{align}

which after simplifications gives

\begin{align} \cos\theta&= \sin\tfrac\alpha2\left(\tfrac{d}a\cos\tfrac\alpha2 +\sqrt{1+(\tfrac{d}a\cos\tfrac\alpha2)^2}\right) . \end{align}

Given that, it is straightforward to find the other two points of the triangle $ABC$.

Edit

For example, like this:

Draw the line $\mathcal{L}\perp GH$ through the point $G$.

Draw the circle $\mathcal{C}$ with the center at $D$ and the radius $r$.

The point $P=\mathcal{L}\cap \mathcal{C}$.

The point $B=PD \cap AH$.

The point $C=PD \cap AG$.

Where the radius is

\begin{align} r=|PD|&= \frac{g}{\cos\theta} =\frac{d\tan\tfrac\alpha2}{\cos\theta} \\ &= \frac{a}{\cos\tfrac\alpha2\left(\cos\tfrac\alpha2 +\sqrt{(\tfrac{a}d)^2+\cos^2\tfrac\alpha2}\right)} . \end{align}