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.
Prove: (2)
which is equivalent to
That is, PC and QC are half the harmonic mean of the legs of right triangle ABC.
Prove: (3)
The altitude CK is concurrent with AD and BG at H.
Hint: Use Ceva's Theorem.
Prove: (4)
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.
.
Prove: (5)
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.
Reference:
Charosh, M. (1965) Mathematical Challenges Washington, DC: National Counctil of Teachers of Mathematics.