A line through a vertex of a triangle cuts it in two similar triangles with √3 as the ratio between correspondent sides.
Find the angles of the given triangle.
Assume the cevian is not an altitude. Then two different angles are formed where it meets the side. If one angle has measure T degrees, then the other measures 180-T degrees. Because we are assuming this is not an altitude then T and 180-T are different angles.
But then for the smaller triangles to be similar, those triangles must have an angle of T and an angle of 180-T and a third angle.
That sum exceeds 180 degrees, so is impossible.
Therefore the cevian is an internal altitude. There are two cases where an internal altitude partitions a triangle into two smaller similar triangles. It is either it is the altitude of an isosceles triangle or the altitude of a right triangle.
The isosceles triangle case: The altitude partitions the triangle into two congruent triangles, which by definition have a ratio of 1, not sqrt(3). This case then must be discarded.
The right triangle case: The internal altitude partitions the triangle into two smaller triangles, and all three triangles are similar but none are congruent. (Except when the triangle is the isosceles right triangle, but that is already discarded in the first case.)
To be more precise let ABC be the main right triangle with right angle at C. Let CD be the internal altitude. Then ABC, ACD and CBD are similar triangles.
Also CD is the geometric mean of AD and DB.
Wlog, AD<DB. Then AD*sqrt(3)=CD.
AD and CD are the legs of ACD. Then sqrt(3) = CD/AD = tan(angle CAD).
Then angle CAD = arctan(sqrt(3)) = 60 degrees. All the triangles are 30-60-90 right triangles.
Solution
