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Solution to puzzle 101: Right triangles

triangleABC is right-angled at A.  D is a point on AB such that CD = 1.  AE is the altitude from A to BC.  If BD = BE = 1, what is the length of AD?


As with puzzle 100, several approaches are possible.  Of the two solutions below, the first, though longer, is more elementary.

Geometric Solution

Let AD = x, CE = y, and angleABC = t.  Let AE and CD meet at F.
Since triangleBCD is isosceles, angleBCD = t.
Hence angleCFE = 90° − t, and so angleDFA= 90° − t.
Since also angleFAD = angleEAB = 90° − t, triangleDFA is isosceles, and so DF = AD = x.
Hence CF = 1 − x.

Triangle ABC, with cevians as described above. Let AD=x, CE=y, angle ABC=t. Let AE and CD meet at F.

Triangles ABE and CFE are similar, as each contains a right angle, and angleABC = angleECF.
Hence y/(1 − x) = 1/(1 + x), and so
y = (1 − x)/(1 + x)(1)

Triangles ABC and ABE are similar, as each contains a right angle, and angleABC = angleABE.
Hence (1 + x)/(1 + y) = 1/(1 + x), and so (1 + x)2 = 1 + y.

Substituting for y from (1), we obtain (1 + x)2 = 1 + (1 − x)/(1 + x) = 2/(1 + x).
Hence (1 + x)3 = 2.

Therefore the length of AD is cube root of 2 − 1.

Trigonometric Solution

Let AD = x, and angleABC = t.
Since triangleBCD is isosceles, angleBCD = t.
We also have angleBCA = 90° − t, and so angleDCA = 90° − 2t.
Hence angleADC = 2t.

Triangle ABC, with cevians as described above. Let AD=x, and angle ABC = t.

Considering triangles ABE and ADC, we obtain, respectively
cos t = 1/(1 + x)
cos 2t = x

Applying double-angle formula cos 2t = 2cos2t − 1, we get
x = 2/(1 + x)2 − 1

Hence (1 + x) = 2/(1 + x)2, from which (1 + x)3 = 2.

Therefore the length of AD is cube root of 2 − 1.

Source: Mathematical Diamonds, by Ross Honsberger. See More Challenges.

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