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In the below triangle, we are looking for the value of angle $φ$.

We are given $α=30, β=18, γ=24$ and also that $CD=BD$.

I have solved it with trigonometry (sine law) and found the required angle to be 78 but I need to solve it with Geometry only.

enter image description here

What I have tried so far:

First of all, the angle is constructible, which means to me that there must be a geometrical solution. I first drew triangle ABC; easy, since we know 2 of its angles. We are not interested in the lengths of the sides. Then, with side AC as a base, and angle of 24 degrees, we can draw a ray from the point A.

Then, since $CD=BD$, triangle DCB is isosceles, therefore D must lie on the perpendicular bisector of CB, which we can draw. The point of intersection of the ray from A and the perpendicular bisector, is point D.

From triangle FEB we have that

angle AFD = 108.

From triangle AFD,

$ADC+CDE+54+108=180$ so $ADC+CDE=18$

We also have $24+ACD+ADC=180$

$ACB=132$

$132+φ+ACD=180$

$18+φ+54+ADC+2CDE=180$

I am always one equation short.

Any ideas?

Many thanks in anticipation!

EDIT:

Sine law in triangle ABD:

$\frac {sin (φ+18)}{AD} = \frac {sin (54)}{BD}$

Sine law in triangle ACD:

$\frac {sin (360-132-φ)}{AD} = \frac {sin (24)}{CD} = \frac {sin (24)}{BD}$

so

$\frac {sin (φ+18)}{sin (228-φ)} = \frac {sin (54)}{sin (24)}$

hence $φ=78$.

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    $\begingroup$ Can you present your sine rule solution? It's unlikely that a pure angle chasing approach (which seems to be all that you currently have) will suffice. $\endgroup$ – Calvin Lin Jan 7 at 14:42
  • $\begingroup$ I don't think it has a single solution $\endgroup$ – Aven Desta Jan 7 at 14:59
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    $\begingroup$ @CalvinLin I don't want to use trigonometry in the solution. I only used it to somehow be guided to the geometrical solution. Also Geogebra shows that there is a unique solution $φ=78$. I will edit my post to present my trigonometrical approach. $\endgroup$ – Nhung Huyen Jan 7 at 15:04
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    $\begingroup$ I think that if we choose a different value for $\alpha$, $\beta$, or $\gamma$, then the angle we want could be made non-constructable. This shows that pure angle chasing won't be enough. It also suggests that there must be something about each particular angle value that must be used, e.g. that $\alpha$ is half the vertex angle of a regular triangle and that $\beta$ is half the central angle of a regular decagon. $\endgroup$ – Jaap Scherphuis Jan 7 at 15:53
  • $\begingroup$ These angles are so off! Usually I just want to reflect stuff and see how many angles I can get to make like a $60$ so I can get equilateral angles to work with, like here : math.stackexchange.com/questions/2613282/… See if this helps! $\endgroup$ – Teresa Lisbon Jan 7 at 15:54
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Consider a regular $30$-gon $X_1X_2X_3X_4X_5X_6X_7X_8X_9X_{10}X_{11}X_{12}X_{13}X_{14}X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}X_{22}X_{23}X_{24}X_{25}X_{26}X_{27}X_{28}X_{29}X_{30}$ and place it on the plane so that $X_1 \equiv A$, $X_6\equiv B$, and that $X_2$ and $C$ lie on different halfplanes determined by the line $AB$. Denote $K=X_2$, $L=X_3$, $M=X_4$, $N=X_5$, and $X_{15}=R$.

Build regular pentagon $KLOPQ$ as in the picture. We shall prove that $P\equiv C$.

Note that $\angle QKA = \angle LKA - \angle LKQ = 168^\circ - 108^\circ = 60^\circ$. Since $QK=KL=AK$, it follows that the triangle $AKQ$ is equilateral. In particular, $AQ=KQ=QP$, so $Q$ is the circumcenter of $AKP$. Angle chasing yields $\angle AQP = 360^\circ - 2\angle PKA = 360^\circ - 2(60^\circ + 36^\circ) = 168^\circ$, so by SAS triangle $AQP$ is congruent to $KLM$, $MNB$, and by symmetry it is congruent to $MOP$. Continuing angle chasing, $\angle PAQ = 6^\circ$, and finally $\angle BAP = \angle KAQ - \angle PAQ - \angle KAB = 60^\circ - 6^\circ - 24^\circ = 30^\circ$.

On the other hand, by congruency of $KLM$, $MNB$ and $MOP$, we have $MK=MP=MB$, so $M$ is the circumcenter of $KPB$ and therefore $\angle BMP = 2\angle BKP = 2(\angle LKP - \angle LKB) = 2(72^\circ - 18^\circ) = 108^\circ$, hence $\angle PBM = 36^\circ$ and $\angle PBA = \angle PBM - \angle ABM = 36^\circ - 18^\circ = 18^\circ$.

Since $\angle BAP = 30^\circ$ and $\angle PBA = 18^\circ$, we have that $P\equiv C$.

We shall prove now that $R\equiv D$. First of all, we have $\angle CAR = \angle BAR - \angle BAC = 54^\circ - 30^\circ = 24^\circ$. Secondly, since $\angle LKC = 72^\circ = \angle LKR$, we have that $K$, $C$, $R$ are collinear. Since $M$ is the circumcenter of $CKB$, we have $\angle BCR = \frac 12 \angle BMK = \frac 12 \cdot 156^\circ = 78^\circ$. We also have $\angle RBC = \angle RBA - \angle CBA = 96^\circ - 18^\circ = 78^\circ$. Since $\angle BCR = \angle RBC$, it follows that $R$ lies on the perpendicular bisector of $CB$, which along with $\angle CAR = 24^\circ$ means that $R\equiv D$. The answer follows: $$\varphi = \angle BCD = \angle BCR = 78^\circ.$$

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    $\begingroup$ This is a whole NEW level , +1, I feel like it is a viable strategy, when you have angles that are multiples of $6, 10$ or some common number that is a divisor of $360$, then you can kind of embed the whole setup into this "n" gonal situation, and suddenly you have so many relations to play with you are spoilt for choice. $\endgroup$ – Teresa Lisbon Jan 8 at 17:58
  • $\begingroup$ I would just complete the proof with an addition (may be starting the proof with this) that shows that there is only ONE solution. This may be achieved by simple geometric means. $\endgroup$ – Moti Jan 10 at 4:04
  • $\begingroup$ Your approach is GREAT. The proof may be simplified by using the fact that there is only one solution and than just pick the right segments to get the desired triangles by connecting vertices of the 30-gon structure. $\endgroup$ – Moti Jan 10 at 4:08
  • $\begingroup$ what tools did you use to make the image/diagram? $\endgroup$ – Aven Desta Jan 10 at 9:03
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    $\begingroup$ @AvenDesta Geogebra $\endgroup$ – timon92 Jan 10 at 9:17
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Since $\angle DAB=54^o$, if we construct a regular pentagon on $AD$, then $AB$ bisects $\angle DAG=108^o$, and $AB$ extended to $K$ on the circumcircle passes through center $N$.

Extend $AC$ to $I$, $DB$ to $L$, and join $IK$, $KL$, $LA$, $IL$, and $DG$.

Since cyclic quadrilateral $AIKL$ has a right angle at $I$, it is a rectangle. Therefore $\angle AIL=\angle IAK=30^o$, $\angle LAK=60^o$, and$$\angle LAG=\angle LAK-\angle GAK=60^o-54^o=6^o=\angle LDG$$And since in the regular pentagon $\angle ADG=36^o$, and as OP notes $\angle ADE=18^o$, then $\angle LDG=\angle ADC$.

size of angle phi Therefore$$\angle CDB=\angle ADG-2\angle LDG=36^o-2\cdot6^o=24^o$$and$$\angle DCB=\phi=\frac{180^o-24^o}{2}=78^o$$

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    $\begingroup$ But how can the segment IN, when extended, necessarily meet the circle at point L? $\endgroup$ – DR SK MOBINUL HAQUE Jan 11 at 14:01
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    $\begingroup$ You are right of course. $AIKL$ is a rectangle only if $IL$ passes through $N$, making triangle $ANI$ isosceles, $\angle AIL=30^o$, $LAK=60^o$, etc. My argument relies on an unproven assumption. Right angles are needed at $A$ and $K$, not just at $I$ and $L$, in order to have the rectangle, and for this $IL$ must go through $N$. I considered deleting the post, but maybe someone can see a way to repair it? $\endgroup$ – Edward Porcella Jan 11 at 16:39

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