Find all angles in a triangle, given 2 internal 90deg angles and segment equality.

Question

Given the triangle ABC, with $\measuredangle ABC > 90^\circ $, points D and E are on AC, such that [AD] = [DE] = [EC] and $\measuredangle ABE = 90^\circ $ and $\measuredangle DBC = 90^\circ $.

Find the measures of the angles in triangle ABC.

(Source: Romanian Math Magazine (Gazeta Matematica)).

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I tried denoting angles BAC and BCA with x and y, but I seem to be going in circles. Also tried to construct the height from B, and use the height theorem for right angled triangles, without getting to the result.

Answer 1

$BD$ is a median in the right triangle $ABE$, hence $BD = AD = DE$; $BE$ is a median in the right triangle $DBC$, hence $BE = DE = EC$. Therefore $BDE$ is an equilateral triangle.

Can you take it from here?

Answer 2

Found an alternate solution, for those not knowing the theorem :)

Construct EF perpendicular on BC; then, triangles ECF and DCB are similar (not sure the English term for triangles with proportional edges), becaue EF || BD; so, F is the middle of BC; EF is both median and height => BE = EC.

Then, follow the same logic to show BDE is equilateral.