$\triangle ABC$ with a point $D$ inside has $\angle BAD=114^\circ$, $\angle DAC=6^\circ$, $\angle ACD=12^\circ$, and $\angle DCB=18^\circ$.

Question

Let $ABC$ be a triangle with a point $D$ inside. Suppose that $\angle BAD=114^\circ$, $\angle DAC=6^\circ$, $\angle ACD=12^\circ$ and $\angle DCB=18^\circ$. Show that $$\frac{BD}{AB}=\sqrt2.$$

I am requesting a geometric proof (with as little trigonometry as possible). A completely geometric proof would be most appreciated. I have a trigonometric proof below.

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Trigonometric Proof

Wlog, let $AB=1$. Note that $\angle ABC=\angle ACB=30^\circ$, so $AC=1$. Then by law of sines on $\triangle ACD$, $$AD=\frac{\sin 12^\circ}{\sin 18^\circ}.$$ By law of cosines on $\triangle ABD$, $$BD^2=1^2+\frac{\sin^212^\circ}{\sin^2{18^\circ}}-2\frac{\sin 12^\circ}{\sin 18^\circ}\cos 114^\circ.$$ As $\cos 114^\circ=-\sin24^\circ$, we get $$BD^2=2+\frac{-\sin^218^\circ+\sin^212^\circ+2\sin12^\circ\sin18^\circ\sin 24^\circ}{\sin^218^\circ}.$$ Then from the identities $\sin^2\alpha-\sin^2\beta=\sin(\alpha-\beta)\sin(\alpha+\beta)$ and $\sin(2\alpha)=2\sin\alpha\cos\alpha$, we have $$BD^2=2+\frac{-\sin 6^\circ\sin 30^\circ+4\sin 6^\circ\cos 6^\circ \sin 18^\circ\sin24^\circ}{\sin^218^\circ}.$$ Because $\sin 30^\circ=\frac12$, we conclude that $BD=\sqrt{2}$ if we can prove $$8\cos 6^\circ \sin 18^\circ \sin 24^\circ=1.$$ This is true because by the identity $2\sin\alpha\cos\beta=\sin({\alpha+\beta})+\sin(\alpha-\beta)$, we have $$2\sin 24^\circ \cos 6^\circ =\sin 30^\circ+\sin 18^\circ.$$ Since $\sin 30^\circ=\frac12$, we obtain $$8\cos 6^\circ \sin 18^\circ \sin 24^\circ =2\sin 18^\circ +4\sin^218^\circ=1,$$ noting that $\sin 18^\circ=\frac{\sqrt5-1}{4}$.

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Attempt at Geometric Proof

I discovered something that might be useful. Construct the points $E$ and $G$ outside $\triangle ABC$ so that $\triangle EBA$ and $\triangle GAC$ are similar to $\triangle ABC$ (see the figure below). Clearly, $EAG$ is a straight line parallel to $BC$. Let $F$ and $H$ be the points corresponding to $D$ in $\triangle EBA$ and $\triangle GAC$, respectively (that is, $\angle FAB=\angle DCB=\angle HCA$ and $\angle FAE=\angle DCA=\angle HCG$). Then $\triangle FBD$ and $\triangle HDC$ are isosceles triangles similar to $\triangle ABC$, and $\square AFDH$ is a parallelogram. I haven't been able to do anything further than this without trigonometry.

Here is a bit more attempt. If $M$ is the reflection of $A$ wrt $BC$, then through the use of trigonometric version of Ceva's thm, I can prove that $\angle AMD=42^\circ$ and $\angle CMD=18^\circ$. Not sure how to prove this with just geometry. But this result may be useful. (Although we can use law of sines on $\triangle MCD$ to get $MD$ and then use law of cosines on $\triangle BMD$ to get $BD$ in terms of $AB$ too. But this is still a heavily trigonometric solution, even if the algebra is less complicated than the one I wrote above.)

I have a few more observations. They may be useless. Let $D'$ be the point obtained by reflecting $D$ across the perpendicular bisector of $BC$. Draw a regular pentagon $ADKK'D'$. Geogebra tells me that $\angle ABK=54^\circ$ and $\angle AKB=48^\circ$. This can be proven using trigonometry, although a geometric proof should exist. But it is easy to show that $KD\perp CD$ and $K'D'\perp BD'$.

In all of my attempts, I always ended up with one of the following two trigonometric identities: $$\cos 6^\circ \sin 18^\circ \sin 24^\circ=1/8,$$ $$\cos 36^\circ-\sin18^\circ =1/2.$$ (Of course these identities are equivalent.) I think a geometric proof will need an appearance of a regular pentagon and probably an equilateral triangle, and maybe a square.

Answer 1

Let $\omega$, $O$ be the circumcircle and circumcenter of $\triangle ABC$, respectively. Let $P,Q,R,S$ be four points on the shorter arc $AC$ of $\omega$ dividing this arc into five equal parts.

First, we shall prove that $\triangle RSD$ is equilateral. Let $D'$ be a point inside $\omega$ such that $\triangle RSD'$ is equilateral. Also, let $E$ be inside $\omega$ such that $\triangle PQE$ is equilateral. Invoking symmetries we see that $\triangle D'SC \equiv \triangle D'RQ \equiv \triangle EQR \equiv \triangle EPA$. Note that $\angle EQR = \angle QRD'=\angle QRS-60^\circ = 168^\circ - 60^\circ = 108^\circ$. Hence $\angle D'QR = 90^\circ - \frac 12\angle QRD' = 36^\circ$ and $\angle EQD'=108^\circ - 36^\circ = 72^\circ$. But also $\angle D'EQ = 180^\circ - \angle EQR = 180^\circ - 108^\circ = 72^\circ$. Hence $ED'Q$ is isosceles with $QD'=ED'$. Again, using symmetries we see that $AED'C$ is an isosceles trapezoid with $AE=ED'=D'C$. We have $\angle ACD'=\angle SCD' - \angle SCA = 36^\circ - 24^\circ = 12^\circ$. Since $AED'C$ is an isosceles trapezoid, it is cyclic and since $AE=ED'=D'C$, it follows that $\angle D'AC = \frac 12 \angle EAC = \frac 12 \angle ACD'=6^\circ$. Hence $D'$ coincides with $D$.

Now comes my favourite part. Some angle chasing shows that $\angle QCE = 18^\circ = \angle DCB$ and $\angle DQC = 24^\circ = \angle BQE$. Hence $D$ and $E$ are isogonal conjugates in $\triangle BQC$. It follows that $\angle CBD = \angle EBQ$.

Choose $T$ on $\omega$ so that $BT$ is a diameter. Clearly, $\triangle BQE$ is symmetric to $\triangle TRD$ with respect to perpendicular bisector of $QR$. In particular, $\angle RTD = \angle EBQ$.

Let $RT$ intersect $BC$ at $X$. Since $\angle CBD = \angle EBQ = \angle RTD$, quadrilateral $BDXT$ is cyclic. Hence $\angle BDT = \angle BXT$. Then some angle chasing shows that $\angle DOB = 102^\circ = \angle BXT = \angle BDT$. This precisely means that the circumcircle of $DOT$ is tangent to $BD$ at $D$. Tangent-secant theorem yields $BD^2=BO\cdot BT = BO \cdot 2BO = 2BO^2$. Hence $$\frac{BD}{AB} = \frac{BD}{BO} = \sqrt 2,$$ as desired.

Answer 2

This is an in-complete proof because I am stuck from step 8 onward.

1. Draw Z on BC such that $\angle BAZ = 90^0$.

2. Let CZ = 1. Then AZ = 1 because $\triangle ZAC$ is isosceles.

3. Because $\angle ABC = 30^0$, $AC = AB = \sqrt 3$.

4. Construct the blue circle (centered at B, radius $= BA = \sqrt 3$.

5. From D draw the tangent to circle (B) touching it at X. Then $\angle BXD = 90^0$,

6. Let Y be the midpoint of BC.

7. Draw CV // YX. By intercept theorem, BX = XV. Together with the finding in (5), we can say that DX is the perpendicular bisector of BV.

8. Draw the circle passing through B, D, V. It will cut the red circle (A) at some point U. [Another way to let X be the center of the dotted circle and prove that D is a con-cyclic point of that circle.]

If we can show that X is the center of the dotted circle, then $\triangle DBV$ is 45-45-90. In turn, $\triangle XBD$ is also 45-45-90. Consequently, required result follows from the fact that $BD = \sqrt 2 \times \sqrt 3$.

Answer 3

New version

1. Draw Z on BC such that $\angle BAZ = 90^0$. If we let CZ = 1, then AZ = 1 because $\triangle ZAC$ is isosceles. Further, because $\triangle ABZ$ is 30-60-90, $BZ = 2$, and $AC = AB = \sqrt 3$.

2. Draw the red circle (A) with radius $= AB = \sqrt 3$. It will cut AZ extended at W. Note that $\angle WBC = 0.5 \angle WAC = 15^0$. Therefore, $\triangle ABW$ is 45-45-90.

3. Extend WD to cut circle (A) at U.

4. Construct the blue circle (B) with radius $= BA = \sqrt 3$. From D draw the tangent to circle (B) touching it at X. Let the tangents ZA and DX meet at T. By tangent properties, $\angle BSA = 90^0$. Then, it is not difficult to prove that XBAU is a rhombus.

4.5 [Added: BU is a chord of circle (A). Let S be the midpoint of BU. Then, $\angle ASB = 90^0$.

In the circum-circle XBAT, its diameter (BT), will perpendicularly bisect AX.

The two together imply BSTU is a straight line.]

5. Let the dotted circle passing through B, D, and U cut BX extended at V. Then, $\angle BVD = \angle BUD = 0.5 \times \angle BAW = 45^0$.

6. Since $\angle VXU = 2 \times \angle VBU$ and $\angle VXU = 2 \times VDU$, we can say that X is the center of the dotted circle. Then, $\triangle BDV$ is 45-45-90.

7. In turn, $\triangle BXD$ is also 45-45-90. This means $BD = \sqrt 2 \times BX = \sqrt 2 \times BA$.