3
$\begingroup$

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 $\measuredangle{APB}$?

enter image description here

$\endgroup$
8
  • $\begingroup$ I have added a figure made with Geogebra. $\endgroup$ – Jean Marie Jul 9 '17 at 4:16
  • $\begingroup$ 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? $\endgroup$ – M. Chen Jul 9 '17 at 4:17
  • $\begingroup$ 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. $\endgroup$ – Jean Marie Jul 9 '17 at 4:25
  • $\begingroup$ yeah that make this a lot easier... Thank you! $\endgroup$ – M. Chen Jul 9 '17 at 4:28
  • $\begingroup$ um this is still somehow confusing. Can you give me another hint? Thank you. $\endgroup$ – M. Chen Jul 9 '17 at 4:31
5
$\begingroup$

After some angle chasing, we have the following:- enter image description here

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

Added:-

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

enter image description here

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.

$\endgroup$
6
  • $\begingroup$ um dont it get proved immediately by the subtended arc theorem? $\endgroup$ – M. Chen Jul 9 '17 at 5:24
  • $\begingroup$ @Mick If $C$ is a circumcenter it's holds. But why from the given follows that $C$ is a circumcenter? $\endgroup$ – Michael Rozenberg Jul 9 '17 at 5:29
  • $\begingroup$ @MichaelRozenberg See added. $\endgroup$ – Mick Jul 9 '17 at 5:33
  • $\begingroup$ @Mick Why $C\in AF$? $\endgroup$ – Michael Rozenberg Jul 9 '17 at 5:35
  • $\begingroup$ @MichaelRozenberg F is constructed by extending AC to cut the circumcircle at F. $\endgroup$ – Mick Jul 9 '17 at 5:37
3
$\begingroup$

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 $\measuredangle APB=54^{\circ}+45^{\circ}=99^{\circ}.$

$\endgroup$
3
  • $\begingroup$ I can't follow why KCLA is cyclic after setting CK and CL as (angle) bisectors. $\endgroup$ – Mick Jul 9 '17 at 8:48
  • $\begingroup$ @Mick Because $\measuredangle KCL+\measuredangle KAL=45^{\circ}+135^{\circ}=180^{\circ}$. $\endgroup$ – Michael Rozenberg Jul 9 '17 at 12:26
  • $\begingroup$ Uh! I see. Nice solution using angle bisector theorem. +1 $\endgroup$ – Mick Jul 9 '17 at 17:08
1
$\begingroup$

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∘.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.