On right triangle ABC with right angle at C, construct squares CBDE and ACFG.
Construct AD and BG intersecting at H.
Draw altitude CK through H to AB.
Click here for a GSP Sketch of this configuration.
which is equivalent to
That is, PC and QC are half the harmonic mean of the legs of right triangle ABC.
The altitude CK is concurrent with AD and BG at H.
Hint: Use Ceva's Theorem.
Hint: Use Menelaus's Theorem
Note: This means CH is one half the Harmonic mean of the altitude and the hypothenuse of the triangle ABC.
CH is the length of the side of a square inscribed in triangle ABC with one side lying along AB.
Help See Square Inscribed along a Base of Any Triangle
For any triangle the length s of the inscribed square along a given base is given by the formula
where a is the length of the base and h is the altitude to the the base. Thus in the current problem CH = s, AB = a, and CK = h.
Charosh, M. (1965) Mathematical Challenges Washington, DC: National Counctil of Teachers of Mathematics.