What’s the measure of the $\angle ABD$ in the triangle below

euclidean-geometryplane-geometry

For reference: In triangle $ABC$ the cevian $BD$ and the median $CM$ are drawn; the angle bisector of $\angle BCM$ bisects $BD$ at $N$.
.Calculate $\angle ABD$ if $AD=2, DC= 1$ and $BC=\sqrt3$
(Resolution by geometry if possible – Answer $15^o$)

My progress:
enter image description here

$AM = BM\\BN=DN$

$TH. Menelao:S\triangle AEC-DB:\\: AD.CN.EB = CD.EN.AB \implies 2.CN.EB = EN.AB\\
Th. Menelao:S\triangle AMC-DB:\\AD.CF.MB = CD.FM.AB \implies CF = FM\\
Th. Menelao:\triangle ABD-MC:\\AM.BE.DE=MB.DE.AC \implies BE = 3DE\\
Th. Menelao:\triangle ABD-NC:\\ \implies AN = 3BN$

I'm not seeing any more similarities being applied…

Best Answer

enter image description here

$15^\circ$ is not the correct answer. GeoGebra shows the same.

In $\triangle ABD$, traversal $EC$ intersects $BA$ and $BD$ internally and $AD$ externally. Applying Menelaus's theorem,

$ \displaystyle \frac{AC}{CD} \cdot \frac{DN}{NB} \cdot \frac{BE}{EA} = 1 \implies EA = 3 BE$

Or, $EM + MA = 3 BM - 3 EM$.
As $BM = MA$, $BM =2 EM \implies E$ is midpoint of $BM$.

As $CE$ is also the angle bisector, $BC = CM = \sqrt3$ and $CE \perp AB$

If $BE = x$, applying Pythagoras in $\triangle AEC$ and $\triangle BEC$,

$CE^2 = BC^2 - x^2 = AC^2 - (3x)^2 \implies 3 - x^2 = 9 - 9 x^2$

$ \displaystyle x = \frac{\sqrt3}{2}, AB = 2 \sqrt3$

As $BC = CM = MB$, $\triangle BCM$ is equilateral triangle and $\angle ABC = 60^\circ$.

Now given $AD:DC = BA:BC$, $BD$ is angle bisector of $\angle ABC$.

$\therefore \angle ABD = 30^\circ$

(As a side note, $\triangle ACB$ is right triangle with $\angle ACB = 90^\circ$)

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