I was asked a simple (I thought) question about geometry by a friend. After some time I was able to find a solution using trigonometry and algebra but my friend said there was a way to solve it using constructions and plane geometry.
I have tried many things including internet searches but can’t seem to find a proof.
Here’s the question:
Given a triangle ABC with angle A 60 degrees and angle B 40 degrees, extend side BC through C. Now create triangle DAB such that DC is congruent to AB. What is the angle measure of angle ADB?
This is better in diagram form.
Now by the definition we gave, there should be solutions for acute all acute triangles but will be potentially harder.
The only clue I have is that you can extend AC to Z and create equilateral triangle ABZ to help.
Thanks for reading this question! It’s my first time here so sorry if the formatting and other things are wrong.
Also the answer seems to be 30 degrees.
Best Answer
Extend $AC$ to $O$ to form the equilateral $AOB$ and extend $BO=OE$ to form the circumdiameter of $\triangle{EAB}$, in which $\angle{BEA}=30^\circ$. If $D$ lies on the circumcircle, $\angle{BEA}=\angle{BDA}=30 ^\circ$, and we are done.
To prove that $D$ is on the circumcircle, translate $DC$ by $CO$ to $FO$, where $F$ must be on the circumcircle since $CD=AB=R$. Since $\angle{FOE}=\angle{CBO}$ due to parallel lines, $\angle{FOE}=20^\circ$. If we reflect $\angle{EOF}$ about ${OF}$ to ${FOD'}$ (not shown) we subtend another $20^\circ$, with ${OD'}=R$. Finally, translate ${OD'}$ by ${CO}$ exactly as before, and note that since positions $D$ and $F$ represent arcs $+10$ and $-10$ about the perpendicular bisector of $ACO$ (thin green line) that $DF$ is parallel to $ACO$ and thus $D'$ must arrive at exactly $F$ after this translation, proving that $D'$ coincides with $D$. Hence, $D$ lies on the circumcircle.