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)


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.

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

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Charosh, M. (1965) Mathematical Challenges Washington, DC: National Counctil of Teachers of Mathematics.