# Find the two missing angles in a quadrilateral

This problem originates from a student who came asking for help. After spending some time, we couldn't solve this problem using (Euclidean) geometry alone. We had to resort to trigonometry to solve this. I have since spent many hours on it and can't seem to get anywhere. Here is the problem.

Given:

Quadrilateral $ABCD$

Diagonals $\overline{AC}$ and $\overline{BD}$

$m\angle ABD=19^{\circ}$

$m\angle DBC=57^{\circ}$

$m\angle ACD=30^{\circ}$

$\overline{AB}\cong \overline{CB}$

Find:

$m\angle ADB$.

It is almost immediate that $m\angle BAC \cong m\angle BCA$ and you can easily fill in the following angles, shown in green.

But $x$ and (now added) $y$ are still illusive. I have tried everything from extending lines to drawing parallel lines to looking at the circumcircles and nothing seems to work. The green angles are the ones which we can easily deduce and the missing/blank angles, I don't know.

How can we deduce the value of the missing angle $x$ using only high school geometry?

There is obviously enough "information" here because if nothing else, you can just draw the quadrilateral very carefully and just measure the angle. Physically, the missing angle can only be one value. It is constrained. One should be able to deduce this value with a geometric proof using basic theorems without resorting to advanced theorems or even trigonometry.

The missing values are, just for giggles,

## Clarification

High school geometry, at least in the USA, is "distinct" from trigonometry and it doesn't include law of sines or cosines. High school geometry tries to mimic Euclid and his "Elements" where students memorize some of the definitions and axioms and are forced to mindlessly derive theorems in an excruciatingly mind-numbing manner with cumbersome notation. High school geometry also does not include any "advanced" geometric theorems. The most advanced thing an average student might do is something like constructing a regular hexagon. My questions is, again, can this problem be solved using only the material taught in a typical high school geometry class? If yes, then how? If you believe no, then can you give a convincing argument why trigonometry is necessary?

I am hoping that this is solvable with some basic theorems and doesn't require any advanced theorems. If we allow trigonometry, then the problem is easy.

• Hint: BD is a fixed length (relative to BC). So you have BD, $\angle BDA$ and $AD$. So the triangle $BAD$ is fixed the angles $\angle BDA$ and $\angle BAD$ are fixed. Use the law of cosines to figure out exactly what they must be. – fleablood Apr 20 '18 at 21:18
• It doesn't seem to be solvable without the law of sines or something similar. Because both $x$ and $y$ belond to the same triangle no matter how you see it. – GiaFil7 Apr 20 '18 at 21:21
• "Because both x and y belond to the same triangle no matter how you see it."... which means if you can find out the lengths of the three sides you can figure out what they are. You have all the angles of $\triangle BCD$ so you have $BD$. With law of cosines and $\angle BDA$ and sides $AB$ and $BD$ you can figure out $AD$. Then you have all three sides of $\triangle BAD$ and one angle. That's enough to figure out the other two angles. – fleablood Apr 20 '18 at 21:38
• Oh... sorry. You were saying it's not solvable without. True. At least I don't see it if there is something "special" about those particular angles. But we do have the law of sins and cos. That will determine. I doubt highly that they are 79 and 30. – fleablood Apr 20 '18 at 21:44
• Oh, wait! $41 = \frac 12 82$ so $\sin 82 = \sin (41 + 41)$ so maybe this does work out nicely without a calculator. – fleablood Apr 20 '18 at 21:50

This problem is a particular case of a family of problems with broadly the same solution, so I will post this more general solution and then discuss particular instances of it.

Problem. $$\angle BAC=3\angle CAD$$; $$\angle CBD=30^\circ$$; $$AB=AD$$. What is $$\angle DCA$$?

Solution. Let $$\alpha=\angle CAD$$. $$\triangle BDA$$ is isosceles on base $$BD$$. Therefore $$\angle DBA=\angle ADB=90^\circ-2\alpha$$ and $$\angle CBA=120^\circ-2\alpha$$.

Let $$E$$ be on $$BC$$ such that $$AE=AB$$. Then $$\triangle BEA$$ is isosceles on base $$BE$$. Therefore $$\angle AEB=\angle EBA=120^\circ-2\alpha$$, so $$\angle BAE=4\alpha-60^\circ$$, so $$\angle EAD=60^\circ$$.

Therefore $$\triangle AED$$ is equilateral, so $$\angle EAC=60^\circ-\alpha=\angle ACE$$, so $$\triangle CAE$$ is isosceles on base $$CA$$, i.e. $$CE=AE=DE$$, so $$\triangle CDE$$ is isosceles on base $$CD$$. $$\angle CED=2\alpha$$, so $$\angle DCE=90^\circ-\alpha$$, so $$\angle DCA=30^\circ$$, which solves the problem. Note that $$\angle DCA$$ is independent of $$\alpha$$.

To adapt this to the current problem, relabel from $$ABCD$$ to $$BCDA$$ and specify $$\alpha=19^\circ$$.

If $$\alpha$$ is specified as $$20^\circ$$, and $$\angle DBA$$ as $$50^\circ$$, then the problem is [Langley]. $$AB=AD$$ is easily seen, and the proof proceeds as above. The above proof, but with angles as specified in Langley's problem, is due to J. W. Mercer.

If $$\alpha$$ is specified as $$16^\circ$$, then the problem is that at gogeometry. The point-lettering is the same, but the diagram is flipped.

[Langley] Langley, E. M. "Problem 644." Mathematical Gazette, 11: 173, 1922, according to David Darling

Let L be the perpendicular bisector of AC. It cuts AC at M and will pass through B according to the given. The angles of $\angle ABC$ are divided in sizes as shown. Note that BD is the angle bisector $\angle ABM$.

Let P be a point on AB such that $\angle ACP = 30^0$. The line CP cuts BM at V. Let BVM extended cut CD at U. Note that AUCV is a rhombus with $\triangle CUV$ and $\triangle AUV$ being equilateral.

1) Form the green circle passing through D, A, B. Then, by angles in the same segment, $\angle ADB = \theta$ as shown. (The problem is solved if the value of $\theta$ is found.)

2) Form the red circle (center = B, radius = BC = BA). Let AH be the common chord of the two circles. Then, $\theta = \theta’$. Let CP extended cut the red circle at X. Let CD cut the red circle at Y. Since $\angle ABY = 2 \times \angle ACY = … = 60^0$. This means AXBY is a rhombus with $\triangle XAB$ and $\triangle YAB$ being equilateral.

If XY cuts AB at Z, then XY is the perpendicular bisector of AB. From the fact that all the blue-shaded angles are all equal to $30^0$, we can say that ABHY is also a rhombus with $\theta = 30^0$.

• Your figure is a bit cluttered and also difficult to see. Any chance you could clean it up a bit or at least enlarge it? You could for example remove everything to the right of the point H. – Jens Apr 22 '18 at 0:39
• While you are at it, can you also change the labels to match the problem I posted so that it is easier to follow the proof. – Fixed Point Apr 22 '18 at 1:42
• @Jens Have uploaded a larger picture. Cutting off the right side does not help too much because of the vertical limitation. The picture can be further enlarged slightly by double clicking it. – Mick Apr 22 '18 at 11:32
• @FixedPoint The locations of A, B, C, D are based on an equivalent question posted (numbered as 2736192 and was deleted 2 days ago). Even the orientation is different, the logic remains the same. The idea is:- 1) Move $\angle ADB$ to an equivalent position ($\angle AHB$). 2) Discovering several rhombus(es). The hardest part is to prove XBH is a straight line. I tried not to rely on that requirement. – Mick Apr 22 '18 at 11:42
• Thanks for enlarging the figure. Still can't follow your argument, though. You loose me at "Then, $\angle ADB = \theta$ as shown.". At this point you haven't yet defined $\theta$. Or is that sentence your definition? But then what does that have to do with the green circle? I think you need to break down your proof in smaller digestable pieces. – Jens Apr 22 '18 at 13:43

Let $BC = 1$ unit.

Law of sines:

$\frac {BD}{\sin \angle BCD} = \frac {BC}{\sin \angle BDC}$.

So $BD = \frac {\sin \angle BCD}{\sin \angle BDC}$ units.

Law of cosines

$AD^2 = AB^2 + BD^2 - 2|AB||BD|\cos \angle ABD = 1 + BD^2 -2BD\cos\angle ABD$

So $AD =\sqrt {1 + BD^2 -2BD\cos\angle ABD}$

Law of sines

$\frac {\sin \angle BAD}{BD} = \frac {\sin \angle ADB}{BA=1} = \frac {\sin \angle ABD}{AD}$.

So $\angle BAD = \arcsin ( \frac {\sin \angle ABD}{AD}BD)$

And $\angle ADB = \arcsin (\frac {\sin \angle ABD}{AD})$.

• -1 I specifically said using high school geometry in the spirit of Euclid. Trigonometry, law of sines, and law of cosines are "not allowed". It is easy if we allow these. – Fixed Point Apr 20 '18 at 21:59
• Lots of high schools teach basic trigonometry in first year geometry now. In fact, from what I've seen on this forum and others internet sites I'd say most geometry classes teach the definitions of sin, cos, tan and the law of sines and cosines. (but nothing more). I do know that I did not learn in in my geometry class but I also know I was teaching it in high school geometry 6 years later. – fleablood Apr 20 '18 at 22:12
• $\sin x = \frac {\sin 19}{\sqrt{4\cos^2 41 + 1 - 4\cos 41\cos 19}}=2.$ It is not immediately obvious why $4\cos^2 41 + 1 - 4\cos 41\cos 19 = 4\sin^2 19$ (It is, and I can get there with a lot of algebra, but it is very messy) – Doug M Apr 20 '18 at 22:18
• Yeah... I was mucking about with that but not getting there. $82 = 41 + 41$ and $41 + 19 = 60$ and that's just too much of a coincidence not to be significant. But... I couldn't really work my way out of it. – fleablood Apr 20 '18 at 22:29