How to find missing angles in a quadrilateral

Question

I have a quadrilateral ABCD, with diagonals AC and BD. Given are four angles: ∠DAC = 20°, ∠CAB = 60°, ∠ABD = 50°, and ∠DBC = 30°.

Those are the red angles in the above image.

I need to fill in all the other angles. Most are trivial – the angles in blue – but how do I find ∠BDC and ∠ACD? Their sum is 110, obviously, but I can't figure out how to find the individual angles.

Edit: Note that the red angles are examples; I'm looking for a general solution given any values for these angles that form a convex quadrilateral. (They do if ∠DAC + ∠CAB + ∠ABD < 180° and ∠CAB + ∠ABD + ∠DBC < 180; in that case you can draw triangles ABD and ABC and then quadrilateral ABCD.)

Answer 1

A numeric computation using the Laws of Sines and Cosines yields $\angle ACD=30^{\circ}$ and $\angle BDC=80^{\circ}$.

Let the intersection of $AC$ and $BD$ be $E$ and, wlog, let $AB=1$.

\begin{align} AD &= \frac{\sin 50^{\circ}}{\sin 50^{\circ}} = 1 \\ DE &= AD\frac{\sin 20^{\circ}}{\sin 110^{\circ}} = 0.36397023426620234 \\ BC &= \frac{\sin 60^{\circ}}{\sin 40^{\circ}} = 1.3472963553338608 \\ CE &= BC\frac{\sin 30^{\circ}}{\sin 110^{\circ}} = 0.71688141714205145 \\ CD &= \sqrt{CE^2+DE^2-2\times CE\times DE\times\cos 70^{\circ}} = 0.68404028665133743 \\ \angle ACD &= \arcsin\big(\frac{DE}{CD}\sin 70^{\circ}\big) = 29.999999999999996 \\ \angle BDC &= \arcsin\big(\frac{CE}{CD}\sin 70^{\circ}\big) = 79.999999999999986 \end{align}

Note that $\angle DAB = \angle ABC$. I am not sure how to use this to reach an elegant solution, but I am sure that a non-computation solution will use both this observation and $AB=AD$ in a non-trivial way.

Note also that, post-factum, $CD$ is tangent to the circumcircle of $\triangle ABD$ (because we computed that $\angle ACD=\angle BAD$).

Answer 2

Use sine theorem for $\Delta ABC$ and $\Delta ACD:$ $$\frac{AB}{\sin{40}}=\frac{AC}{\sin{80}}$$ $$\frac{AD}{\sin{ACD}}=\frac{AC}{\sin{(20+ACD)}}$$ Noting $AB=AD$, we can find $mACD=30, mBDC=80$.

Answer 3

If you extend AD and BC to meet at E, this becomes the famous "Langley's adventitious angles" problem. I have seen at least four proofs (other than non-proofs using numerical applications of the law of cosines) deriving angle ACD. A good one is this:

Draw a line meeting AB at A, and meeting BC at F such that angle BAF is $20^\circ$. Thus AB = AF, and angle CAF is $40^\circ$.

Angle ABD and angle ADB are both $50^\circ$ so AD = AB, thus AF = AD. In triangle AFD, sides AF and AD are equal and meet at an angle of $60^\circ$ so triangle AFD is equilateral and AF = FD.

Angle FBC = $80^\circ - 20^\circ - 20^\circ = 40^\circ$. In triangle ACB, angle ACB is $180^\circ - 80^\circ - 60^\circ = 40^\circ$. So triangle AFC is isosceles and CF = AF. Then CF = FD.

That shows that triangle CFD is isosceles and angles FCD and FDC are equal. Angle CFD is $180^\circ$ minus angle AFD minus angle AFB which is $180^\circ - 80^\circ - 60^\circ = 40^\circ$. So angle FDC = angle FCD = $70^\circ$.

Finally since angle ACD plus angle ACF add to $70^\circ$, angle is $30^\circ$ (and angle ADC is $80^\circ$).

Answer 4

You don't need trig to solve this, it's completely doable by basic geometry, although you do have to draw a lot of stuff. Calculating those angles is actually a problem already online known as the "world's hardest easy problem" (problem 2) https://www.duckware.com/tech/worldshardesteasygeometryproblem.html

There's a solution on the site, but I've shortly explained it alone. (the quadrilateral from your problem is ABED on the picture, those sequences of 3 letters are supposed to be angles as I couldnt write them normally in paint) https://i.stack.imgur.com/XMs7Y.jpg

Answer 5

Let $AC\cap DB=\{F\}$ and $L\in FC$ such that $\measuredangle ABL=60^{\circ}$.

Also, let $BL\cap AD=\{G\}$.

Since $\measuredangle DAB=\measuredangle CBA$, we see that $GC||AB$ and from here $GC=GL=LC$.

Now, since $\measuredangle DBA=\measuredangle ADB=50^{\circ}$, we obtain: $AL=AB=AD$, which says that $\measuredangle ADL=\frac{180^{\circ}-20^{\circ}}{2}=80^{\circ}$ and since $\measuredangle AGB=180^{\circ}-80^{\circ}-60^{\circ}=40^{\circ}$,

we obtain: $\measuredangle GLD=40^{\circ}$, which says that $DG=DL$.

But also $GC=CL$, which gives that $DC$ is a bisector of $\angle GCL$.

Thus, $\measuredangle DCA=\frac{1}{2}\cdot60^{\circ}=30^{\circ}$ and the rest is smooth.

Answer 6

I have four different and one general solution to this problem, I hope it will help you.