I need to construct a triangle from given base, obtuse angle adjacent to base and difference of two other sides.
Let us try to analyze the scenario.
We are given base BC, obtuse $\angle\text{ACB}$ adjacent to base BC,
and BD equal to AB minus AC.
We need to construct $\triangle\text{ABC}$.
Now in $\triangle\text{ADC}$, $\text{AD} = \text{AC}$. Hence,
$\angle\text{ACD} = \angle\text{ADC}$.
From the given information, I can draw base BC and $\angle\text{ACB}$.
If
I can find the value of $\angle\text{ACD}$, I can draw it and draw an
arc with center at B and radius equal to BD, and then construct the triangle.
However, I find no way to to calculate $\angle\text{ACD}$.
I understand that,
$\angle\text{ACD} = \angle\text{ADC} = \angle\text{DBC} + \angle\text{DCB}$.
But that is where my thought process stops.
Or may be I completely in the wrong direction.
Any suggestion will be appreciated.
Best Answer
Continue $AC$ to $E$ so that $CE=DB$. Connect $E$ to $B$. Draw $BA$ so that $ABE$ is isosceles, i.e. make base angles $AEB$ and $ABE$ to be equal.