The geometry problem from my previous math puzzle has a nice solution. I particularly like it because it is one of these problems that are pretty hard to solve traditionally - unless you perform the right geometric construction - and at that point, the solution becomes trivial.

We will perform the following trick. From the point D we will draw a segment DE (with E on the longer side - AC in our case). E is chosen such that the angle(ABD) = angle(ADE).

At this point, our puzzle solving strategy is this:
a) Derive as much equations as possible from our new construction, and
b) Eliminate from these euqations any segments involving the point "E".

First, we immediately observe that the triangle BAD is similar with the triangle DAE, therefore:

From here we can extract AE without involving any other segments containing the point "E":

AE = AD^2 / BA (2)

Second, because angle(ADE) + angle(EDC) = angle(ADC) = angle(ABD) + angle(BAD) we obtain: angle(BAD) = angle(EDC) = angle(DAE). In conclusion, the triangle ADC is similar with the triangle DEC, so:

DC/AC = EC/DC = DE/AD (3)

From here, we extract EC, again without involving any other segments containing the point "E":

EC = DC^2 / AC (4)

Third, if we divide the other "unused" parts from equalities from (1) and (3) involving DE (but not involving any other segments containing "E") we obtain: (AD/AB) / (DC/AC) = (DE/BD) / (DE/AD). AD simplifies, and we obtain:

AB/AC = DB/DC (5)

So, we have now three inequalities (2), (4) and (5), where the first two one involve the segments AE and EC.

Now, how we can get rid of "E" in AE and EC? We observe that AC = AE + EC, and by substituting AE and EC from the equations (2) and (4) above, we finally obtain something that doesn't contain any "E"s anymore:

AC = AD^2 / AB + DC^2 / AC. (6)