Construct a triangle, given the measures of two angles and the length of the bisector of the thrid angle.
Draw a triangle with the given angles and side connecting them of unit length (I will call this side the base). Bisect the third angle and draw in the bisector.
Produce the bisector and mark the reqired length from the point that it meets the base through the bisected angle. Now construct line paralle to the sides other than the base through this marked point and you are done.
That is you construct a triangle with the given angles but arbitary size, draw in the bisector, mark the required length on the bisector produced and construct a triangle of the required size similar to the one you drew earlier.
CB
Let θ1 and θ2 are the given angles.
The third angle is θ = 180 - ( θ1 + θ2)
Let PO is the angle bisector. Draw a line PQ such that angle OPQ = θ/2
Mark a point Q' on PQ such that PQ'/PO = sin(θ/2)
Taking OQ' as radius draw a circle. Draw PS tangent to the circle which touches the circle at S'.
Draw line OA such that angle Q'OA = π/2- θ1/2 and OB such that angle S'OB = π/2- θ2/2.
If A and B are the points on PQ and PS, then PAB is the required triangle.
Hello, MATNTRNG!
Here is Captain Black's excellent construction.
Construct a triangle, given the measures of two angles, $\displaystyle \alpha$ and $\displaystyle \beta$
and the length of the bisector of the thrid angle, $\displaystyle b$.
Draw a horizontal line $\displaystyle PQ.$
At $\displaystyle P$ construct angle $\displaystyle \alpha$, at $\displaystyle Q$ construct angle $\displaystyle \beta,$
. . intersecting at $\displaystyle R.$
Code:R o * * * * * * * * * * * * * * * * * α β * P o * * * * * * * * * o Q
Construct the bisector of angle $\displaystyle R$, intersecting $\displaystyle PQ$ at $\displaystyle S.$
On $\displaystyle SR$, measure off $\displaystyle SC = b.$
Code:R o * * * * * * * * * * * * * o C * * * * * b * * α β * P o * * * * o * * * * o Q S
Through $\displaystyle C$, construct a line parallel to $\displaystyle PR,$
. . intersecting $\displaystyle PQ$ at $\displaystyle A.$
Through $\displaystyle C$, construct a line parallel to $\displaystyle QR,$
. . intersecting $\displaystyle PQ$ at $\displaystyle B.$
Code:R o * * * * * * * * * C * * o * * * * * * * * * * α * * β * P o * * o * * * o * * o Q A B
$\displaystyle \Delta ABC$ is the desired triangle.