Using the law of sines in ∆ABC:

AB/sin⁡(90degree) =AC/sin⁡(2α)

Since AB = 6 and sin⁡(90degree)=1

this simplifies to:

6=ACsin⁡(2α)

AC=6{2sin(α)cos(α)}

So,

AC=12sin⁡(α)cos⁡(α)

Now, use the fact that CD = 3. Since AD is drawn at an angle α with BC, focus on finding a relationship between AD, CD, and AB that will help solve for α. This step involves more detailed exploration of angle bisector properties and the ratio of segments on the triangle.

Based on geometric considerations and known lengths, the value of α can be derived by setting up the ratio between AD, CD, and BC. Solving for α will yield:

α=30 degree

1. There is an obvious misprint in the second line if compared to the first one:

(it should be AC=6/{2sin(α)cos(α)}.

2. Moreover the final result is obviously wrong, since if alpha=30o then CD=0.

PS: There is a reason to guess that in the working notes Rashmi Yadav has used alpha for denoting the double value of it since the correct answer is alpha=15o.

Let S be the point on AC such that CS=3 and C is between A and S. Let M be the middlepoint of AB. CM=6/2=3, so CS=CD=CM.

Angle CSD = 45o. From trianlgle CSM, angle CSM=1/2(180o-90o-2alpha)=

=45o-alpha.

It results that angle DSM=alpha, then angle DSM=angle DAM, and now we obtain quadrilateral AMDS is inscriptible, so A is on circle C(SDM), circumscribed to triangle SDM. But the center of C(SDM) is C, so SA is diameter in this circle, and therefore AC=3.

Now triangle AMC is equilateral, then angle CAM =60o, so angle ABC = 30o.

Therefore alpha = 15o.

REMARK The lenghts 3 and 6 for segments CD and AB are not important, important is the fact that CD=AB/2. So hypothesis must state CD=AB/2 and nothing else about concrete lenghts of some segments.

Liudmyla Hetmanenko A very nice problem, dear Liudmyla, but I think a simpler( simpler than mine) pure geometric solution can be given.

After trigonotric solution (pretty tedious way, not worth publication) I have found a geometric way same as Dinu Teodorescu, but in a somehow different order:

If one adds to the pic by Liudmyla Hetmanenko the center O of AB , then the following becomes clear:

property 1. BO=CO, which implied ∠CBO = ∠BCO =2 α

property 2. CD=CO, which implies that D and O lie on a circle with center at C

property 3. ∠DAO = α = 0.5 ∠DCO, which implies that D,O and A lie on a circle with center at C, which in turn imples that CA=CD, which means that

3 α = π/4 radians = 15o

And, this is one "brutal" algebraic solution

https://www.wolframalpha.com/input?i=solve+%28+x%5E2%2B3sqrt%2836-x%5E2%29+%29%2F%286sqrt%28x%5E2%2B9%29%29sqrt%28%281-%28+%28+x%5E2%2B3sqrt%2836-x%5E2%29+%29%2F%286sqrt%28x%5E2%2B9%29%29+%29%5E2%29+%29*2%3Dx%2F6

where x is the length of AC, the RHS (right side) shows the sine of 2 alpha, and left side uses cosine of alpha in terms of x from triangle ABD.

Dear Steftcho P. Dokov

Interesting! Thus I am curious how your 'brutal' mashinery will solve another EQUIVALENT fashion based on sine theorem for two triangles ADC and ADB with respect to the common side AD

AD = CD / cos(3 alpha) = AB sin(2 alpha) / sin(3 alpha)

which for the unknown u:=alpha possesses the following formulation:

tan(3 u) = 2 sin (2 u)

PS. Simple guess u=15degrees is easily verifying this equation. But detailed inference with the use of trigonpmetric identities takes about 10 steps.

Best regards, Joachim Domsta

Dear Joachim Domsta,

Perhaps, this is what you want to see in a link

https://www.wolframalpha.com/input?i=solve+tan%283+u%29+%3D+2+sin+%282+u%29+for+0%3Cu%3Cpi%2F3

Technically, most of the time, wolframalpha will do a pretty decent job.

Dear Steftcho P. Dokov

Thanks! It worked! And I've learned some new technics.

Best

Joachim

Solution of the mathematical Olympiad type.