enter image description here Can you please help me finding the value of angle $x$ in this image (I've drawn using microsoft paint, and added as many angles as many I could figure out). All angles are in degrees. Any exterior angle property or angle sum property seems not to help further.

  • $\begingroup$ What have you tried? What techniques are you supposed to be using? E.g. Are you allowed to use Ceva's Theorem? $\endgroup$ – Calvin Lin Apr 15 at 14:34
  • $\begingroup$ @CalvinLin, this is not a problem I've been given in my institution, neither it is related to my course. I was just trying it for fun, so any technique will do. $\endgroup$ – Martund Apr 15 at 14:35
  • $\begingroup$ Applying the trigonometric form of Ceva's theorem, we get that $ x = 78^\circ$. If you're interested, you can read up on it. (There will be other Euclidean geom / trigo approaches, but this is the fastest.) $\endgroup$ – Calvin Lin Apr 15 at 14:46
  • $\begingroup$ @CalvinLin, thank you, post it as an answer, I'll accept. $\endgroup$ – Martund Apr 15 at 15:06
  • $\begingroup$ I intentionally left it as a comment because I'd like to see an Euclidean geom approach which I think is "nicer". $\endgroup$ – Calvin Lin Apr 15 at 15:07

Draw equilateral triangle $ACE$ such that $E$ lies on the same side of $AC$ as $B$. Then angle chasing shows that $\angle EAB = 18^\circ = \angle BAD$. Since $AC=BC=EC$, we have that $\angle ABE =\frac 12 \angle ACE = 30^\circ = \angle DBA$. Hence triangles $ABE, ABD$ are congruent by ASA. Therefore $AE=AD$, but $AE=AC$, so $AD=AC$. From this we get $\angle ACD = 90^\circ - \frac 12 \angle DAC = 78^\circ$.

Sorry for not marking the angles on the figure. enter image description here

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  • $\begingroup$ Since $AC=BC=EC$, we conclude that $C$ is the circumcircle of $\triangle ABE$. Hence, $\angle BAE=\dfrac12\angle BCE=30^\circ$, which leads to a contradiction apparently. $\endgroup$ – Martund Apr 15 at 17:42
  • $\begingroup$ @Martund You're making a mistake. In fact, $\angle BCE =36^\circ$ and there is no contradiction. $\endgroup$ – timon92 Apr 15 at 17:48
  • $\begingroup$ Thank You, I get it now. What software did you use to draw the image?? $\endgroup$ – Martund Apr 15 at 18:00
  • $\begingroup$ @Martund I used Geogebra. $\endgroup$ – timon92 Apr 15 at 18:04

Draw the angle bisector of $DAC$ and let it meet the $BD$ at $P$.
The rest is just angle chasing and your desire angles are $ACD=78°$ and $BCD=18°$ enter image description here

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  • $\begingroup$ Can you show how you angle chase this? Thanks $\endgroup$ – timon92 Apr 15 at 16:58
  • $\begingroup$ @timon92, We observe that $\angle ABP=\angle BAP=30^\circ$, by construction. Hence, $P$ lies on angle bisector of $AB$ which passes through $C$, since $AC=BC$ is given. By angle sum property of $\triangle ABP$, we get $\angle APB=120^\circ$. By angle sum property of triangle $BPM$, where $M$ is the mid-point of $AB$, we get, $\angle BPM=60^\circ=\angle APM$. Hence, $\angle DPC=180^\circ-\angle BPM=120^\circ$. And$\angle APC=120^\circ$ by sum-of-angles-around-a-point property. $\triangle APD\cong\triangle APC$, by ASA rule. Hence, $DP=CP$. Angle sum property in $\triangle DPC$ will do now. $\endgroup$ – Martund Apr 15 at 17:53
  • $\begingroup$ @Martund Thanks, I see it now. I knew that some congruence should come up somewhere but I could not figure out where. Nice solution. $\endgroup$ – timon92 Apr 15 at 17:57
  • $\begingroup$ @Seyed, what software did you use to draw this figure?? $\endgroup$ – Martund Apr 15 at 18:01
  • $\begingroup$ @Martund, I used C.a.r. (Compass and Ruler) car.rene-grothmann.de/doc_en/index.html $\endgroup$ – Seyed Apr 15 at 19:04

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