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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.

enter image description here

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.

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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.

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  • $\begingroup$ Thanks, looks like I was thinking in the wrong direction. $\endgroup$ – Masroor Mar 1 '13 at 4:51
  • $\begingroup$ AEB will be an obtuse angle.Then there can't be a way to construct an isosceles triangle with another angle equal to AEB. $\endgroup$ – rah4927 Apr 16 '14 at 12:56

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