Solution to puzzle 101: Right triangles

ABC 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 ABC = t. Let AE and CD meet at F.
Since BCD is isosceles, BCD = t.
Hence CFE = 90° − t, and so DFA= 90° − t.
Since also FAD = EAB = 90° − t, DFA 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 ABC = ECF.
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 ABC = ABE.
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 − 1.

Trigonometric Solution

Let AD = x, and ABC = t.
Since BCD is isosceles, BCD = t.
We also have BCA = 90° − t, and so DCA = 90° − 2t.
Hence ADC = 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 − 1.

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