Let $ABCD$ be a convex quadrilateral

with $\measuredangle{ABD} = 18^{\circ}$, $\measuredangle{ACB} = 54^{\circ}$, $\measuredangle{ACD} = 36^{\circ}$ and $\measuredangle{ADB} = 27^{\circ}$.

Let $\{P\}$ be the intersection of the two diagonals $AC$ and $BD$.

What is the degree value of $\measuredangle{APB}$?

• I have added a figure made with Geogebra. – Jean Marie Jul 9 '17 at 4:16
• thank you. This question really confuses me. I am now trying to find sin36 degrees and use law of sine now... Is there a smart way to do this? – M. Chen Jul 9 '17 at 4:17
• A technical remark that will allow you simpler computations is to replace all your angle values by 2a,3a,4a,6a, with a=9°, keeping in mind that 10a=90° for simplifications. – Jean Marie Jul 9 '17 at 4:25
• yeah that make this a lot easier... Thank you! – M. Chen Jul 9 '17 at 4:28
• um this is still somehow confusing. Can you give me another hint? Thank you. – M. Chen Jul 9 '17 at 4:31

After some angle chasing, we have the following:-

The configuration seems like C is the circum-center of the circle DAB.

The tricky part is:- "Is C really the circum-center of $\triangle ABD$?"

This is done by extending AC to cut the circum-circle at F. Again by angle chasing, we get $\beta = \angle 1 = \angle 2$. This means $\triangle CFB$ is isosceles. The same is true for $\triangle CFD$. Then, point C (not the red dot anymore) is indeed the circum-center of $\triangle ABD$ (because (1) CB = CF = CD and (2) B, F, D are points on the circumference of the same circle).

Finally, we have $\angle CDB = \angle CBD = 45^0$. Result follows.

• um dont it get proved immediately by the subtended arc theorem? – M. Chen Jul 9 '17 at 5:24
• @Mick If $C$ is a circumcenter it's holds. But why from the given follows that $C$ is a circumcenter? – Michael Rozenberg Jul 9 '17 at 5:29
• @MichaelRozenberg See added. – Mick Jul 9 '17 at 5:33
• @Mick Why $C\in AF$? – Michael Rozenberg Jul 9 '17 at 5:35
• @MichaelRozenberg F is constructed by extending AC to cut the circumcircle at F. – Mick Jul 9 '17 at 5:37

Let $CK$ be a bisector of $\Delta ADC$ and $CL$ be a bisector of $\Delta ABC$.

Thus, $KCLA$ is cyclic and from here $\measuredangle KLA=\measuredangle KCA=18^{\circ}=\measuredangle ABD$.

Thus, $KL||DB$, which says $AC:DC=AK:KD=AL:LB=AC:BC$,

which gives $DC=BC$, $\measuredangle CBA=45^{\circ}$ and $\measuredangle APB=54^{\circ}+45^{\circ}=99^{\circ}.$

• I can't follow why KCLA is cyclic after setting CK and CL as (angle) bisectors. – Mick Jul 9 '17 at 8:48
• @Mick Because $\measuredangle KCL+\measuredangle KAL=45^{\circ}+135^{\circ}=180^{\circ}$. – Michael Rozenberg Jul 9 '17 at 12:26
• Uh! I see. Nice solution using angle bisector theorem. +1 – Mick Jul 9 '17 at 17:08

Let S be the center of the circumcircle of triangle ABD. D is a point on that circumcircle, and ∠ADB = 18∘, so therefore, ∠ASD = 36∘. Consider the circumcircle of ASD. S is a point on that circumcircle and ∠ASD = 36∘, so therefore for each point on that circumcircle Q and only for those points, ∠AQD = 36∘ is a true statement.

We know ∠ACD = 36∘, so therefore, C is on the circumcircle of ADS.

Analogically, since ∠BCA = ∠BSA = 2 * ∠BDA = 54∘, C is also on the circumcircle of ABS.

Out of this, we can conclude points C and S are on the intersections of ABS's and ADS's circumcircles. The only such intersections are points A and C. Obviously, S is not the same point as A (otherwise, ∠BDA = ∠DBA would be true), and neither is C (otherwise, ∠ACB = 54∘ makes no sense) so we conclude S is C.

Therefore, C is ABD's circumcenter. Out of that, we can get DC = BC (two radii in the same circle), and by that we conclude BCD is bilateral and ∠DBC = ∠BDC.

We know ∠BCD = 36∘ + 54∘ = 90∘. Therefore , ∠DBC=45∘.

Finally, ∠APB = 54∘ + 45∘ = 99∘.