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Let AGB be a semicircle and AC = a, BC = b.
If CG ⊥ AB and CH ⊥ OG then GH = H(a, b).
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In a trapezoid ABCD, let EF be the line parallel to the bases AB and CD through the point of intersection of the diagonals AC and BD:
If CD = a and AB = b, then EF = H(a, b).
It is worth observing that the point of intersection of the diagonals divides EF into two equal parts, which leads to another common example:
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In a 1877 sangaku from the Hyogo prefecture, AE and BD are both perpendicular to AB. P is the intersection of AD and BE, and CP ⊥ AB.
If AE = a and BD = b, then CP = 2·H(a, b).
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The same motif reemerges in the See-Saw Lemma where AE, BF, and EF are tangent (the latter at X) to the semicircle AXB.
If AE = a and BF = b, then XY = H(a, b).
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Vladimir Nikolin from Serbia has observed appearance of the harmonic mean in his Rhombus Lemma.
Let AD be the bisector of ∠A in ΔABC, B'D||AC and C'D||AB. Then AB'DC' is a rhombus and its side p is half of the harmonic mean of sides b = AC and b = AB.
Choosing the angle at A to be right leads to a special case where the rhombus becomes a square.
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Let a square be inscribed into a right triangle as in the diagram
Then, if a and b are the legs of the triangle and p the side of the square, 1/p = 1/a + 1/b. In other words, p = H(a, b)/2.
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As a consequence of the Rhombus Lemma there is a simple way of constructing the harmonic ratio:
In ΔABC, with AB = c and AC = b, draw AD - the bisector of ∠A. At D erect a perpendicular to AD and let E be its intersection with either AB or AC. Then AE = H(b, c).
