# Finding an angle between side and a segment from specified point inside an equilateral triangle

Here is the question: • $\overset{\Delta}{ABC}$ is an equilateral triangle.
• D is a point inside triangle.
• $m(\widehat{BAD})=12^\circ$
• $m(\widehat{DBA})=6^\circ$
• $m(\widehat{ACD})=x=?$

I managed to reduce answer to $\cot x=\csc 48^\circ+\sqrt 3$ with the help of Trig-Ceva Theorem, but i can't find x without a calculator. I suspect that there is an elementary (but complex) method to solve this but i'm not sure.

[The answer i'm trying to reach is $18^\circ$]

• Can leave result as $\arctan$? $1/\sin 48^o$ has exact value $8[\sqrt{2}\sqrt{5+\sqrt{5}}+\sqrt{3}(\sqrt{5}-1) ]^{-1/2}$. – Milly Oct 28 '14 at 19:17
• @Milly No. I need to reach an answer which is exactly $18^\circ$. I forgot to wrote that. – Alistair Oct 28 '14 at 19:24
• You might draw a line from C to the middle of AB and construct the mirror of CD (C´D´) to the right of the line. You then need to prove that the continuation of AD meets C´D´ in a right angle. The left half of the top angle is now 18 + 12 degrees. There are a fair amount of small triangles with 12-90-78 degrees you need to work with. It is facilitated be your knowing of the answer. – Mikael Jensen Oct 30 '14 at 23:30
• @MikaelJensen Actually; i try to prove something like you mentioned on left side of the line by drawing segment from $B$ to $[CD]$ such that $m(\widehat{EBD})=6^\circ$ ($E$ is intersection point of segment and $[CD]$.) and showing that segment and $[CD]$ is perpendicular, to no avail. As a result of cannot finding answer with the 'visual' methods, trigonometry followed. If i can just prove the identity $\cot 18^\circ-\sec 42^\circ=\sqrt{3}$ then i'm done. – Alistair Oct 31 '14 at 0:57

This problem is actually part of a regular $30$-gon, as shown in the following figure. For a proof that three diagonal lines meet at a single point, refer to Section $3$ and Table $4$ (page $13$) of http://www-math.mit.edu/%7Epoonen/papers/ngon.pdf. (Unfortunately, this is not so easy. The following lemma from Section $2$ (pages $4-5$) helps.)

Lemma. Let $A$, $B$, $C$, $D$, $E$, $F$ be six distinct points in order on a circle. The three chords $AD$, $BE$, $CF$ meet at a single point if and only if $AF\cdot BC\cdot DE=CD\cdot EF\cdot AB$. • Never thought about changing the problem to polygon diagonals. This nice method can generalize really well. However there are some issues: First; i'm stuck at proving product of ratios of adjacent diagonals is $1$. Peter Steinbach's diagonal product formula can easily reduce this issue to showing $d_3+d_5=d_9$, where $d_i$ is $i$th smallest diagonal. Do you know how i prove this result? ... – Alistair Nov 14 '14 at 18:12
• ... Second; According to Modern geometry; an elementary treatise on the geometry of the triangle and the circle by Roger A. Johnson, your lemma might not be true. According to theorem 228 in page 151, this lemma cannot be used to prove chords are concurrent. I was able to prove concurrent segments imply above relationship easily by Ceva's method, but i'm not sure inverse of this is true, which is needed for solution. Do you know of any proof of this lemma? Where did you encounter this? – Alistair Nov 14 '14 at 18:24
• A proof of $d_3+d_5=d_9$: it suffices to show $\sin 54^{\circ}-\sin 18^{\circ}=\sin 30^{\circ}$, that is $2\cos 36^{\circ}\sin 18^{\circ}=1/2$. Using the fact that $\alpha=\sin 18^{\circ}$ satisfies $4(1-\alpha^2)-3=2\alpha$ (from $\cos 54^{\circ}=\sin 36^{\circ}$), we can show that $2(1-2\alpha^2)\alpha=1/2$. – pharmine Nov 14 '14 at 22:13
• I used $\cos 3\theta=4\cos^3\theta-3\cos\theta$ and $\sin 2\theta=2\sin\theta\cos\theta$. So $\cos 3\theta=\sin 2\theta$ becomes $\cos\theta(4\cos^2\theta-3)=\cos\theta(2\sin\theta)$, and dividing by $\cos\theta$ (which is not zero) we get $4(1-\sin^2\theta)-3=2\sin\theta$. – pharmine Nov 15 '14 at 10:34
• It's simple: note that arc length = angle (in radians) when the radius of the circle is 1 (i.e. unit circle). So when the arc length is $u$, the chord length is $2\sin(u/2)$. – pharmine Nov 18 '14 at 5:28

I document two answers to this question; one is trigonometric answer i came up with, other is geometric answer i found on the net. Geometric answer is in another language, so i translated it to English instead of linking it.

First:

For easier reference, some known formulas:

$$2\sin x\sin y=\cos(x-y)-\cos(x+y)$$

$$2\sin x\cos y=\sin(x+y)+\sin(x-y)$$

$$\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y$$

Second:

Proof that $\sin54^\circ\sin18^\circ=1/4$:

Insert $\sin72^\circ=2\sin36^\circ\sin54^\circ$ into $\sin36^\circ=2\sin18^\circ\sin72^\circ$ to get $\sin54^\circ\sin18^\circ=1/4$.

I solved this question with the Trig-Ceva theorem. From the theorem:

$$\frac{\sin12^\circ}{\sin48^\circ}\frac{\sin54^\circ}{\sin6^\circ}\frac{\sin\alpha}{\sin(60^\circ-\alpha)}=1$$

And we reach:

$$2\cos6^\circ\sin54^\circ\sin\alpha=\sin48^\circ\sin(60^\circ-\alpha)\tag{1}$$

Since $2\sin48^\circ\sin42^\circ=\cos6^\circ$ and $\sin54^\circ\sin18^\circ=1/4$, we can make advance guessing and find $\alpha=18^\circ$ from $(1)$. Else, we need to tamper the equation $(1)$ more:

$$\frac{\sin(60^\circ-\alpha)}{\sin\alpha}=\frac{\sqrt3/2+\sin48^\circ}{\sin48^\circ}\\ \frac{\sqrt3/2\cos\alpha-1/2\sin\alpha}{\sin\alpha}=\sqrt3/2\csc48^\circ+1$$

And we get:

$$\cot\alpha=\csc48^\circ+\sqrt3\tag{2}$$

Now we can prove that $\cot18^\circ=\csc48^\circ+\sqrt3$ by tracing the steps back to equation $(1)$ from equation $(2)$. After that; because we know, since $\cot\alpha$ is one-to-one for $0<\alpha<180^\circ$, there is only one $\alpha$ value to maintain equation $(2)$, we find that $\alpha=18^\circ$. $\blacksquare$

Here is the second proof: Draw angle bisector of $\widehat{BAD}$, intersecting $[BD]$ in $E$. Draw $[CE]$. Since $\overset{\Delta}{ABE}$ is isosceles, $[CE]$ is the axis of symmetry of $\overset{\Delta}{ABC}$, making $m(\widehat{ECB})=30^\circ$. Draw $[BF]$ such that $m(\widehat{FBD})=6^\circ$ and $|BF|=|AB|$. Draw $[AF]$ and $[CF]$. Since $\overset{\Delta}{BCF}$ is isosceles and $m(\widehat{CBF})=48^\circ$, then $m(\widehat{FCB})=66^\circ$, therefore $m(\widehat{FCA})=6^\circ$. Draw $[DF]$. Since $\overset{\Delta}{ABF}$ is isosceles, $[BD]$ is the axis of symmetry of $\overset{\Delta}{ABF}$, making $m(\widehat{DFB})=12^\circ$. Extend $[AD]$; since $[CE]$ is the axis of symmetry of $\overset{\Delta}{ABC}$ and $m(\widehat{FBA})=m(\widehat{BAD})=12^\circ$, extension of $[AD]$ meet at the intersection of two segments $G$. Since $\widehat{FGA}$ is an exterior angle of $\overset{\Delta}{ABG}$, $m(\widehat{FGA})=24^\circ$. Since $\widehat{FDA}$ is an exterior angle of $\overset{\Delta}{DFG}$, $m(\widehat{FDA})=36^\circ$. Since $m(\widehat{FCG})=36^\circ$ opposite angles of $\overset{\square}{CFDG}$ are supplementary, making this a cyclic quadrilateral, meaning its four points are on a circle. Since $\widehat{FGD}$ and $\widehat{FCD}$ both looking at the same arc in this circle, $m(\widehat{FCD})=24^\circ$. And then, finally, we reach $\alpha=18^\circ$. $\blacksquare$