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.

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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.


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.

  • $\begingroup$ 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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