 ## Some Relationships in a Right Triangle

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: (1) Prove: (2) and 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.

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Prove: (5)

CH is the length of the side of a square inscribed in triangle ABC with one side lying along AB.

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.

Special thanks to Jim Metz of Honolulu for calling these problems to my attention, discussing solutions, and raising questions.