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Solution to puzzle 22: Isosceles angle

Skip restatement of puzzle.Let ABC be an isosceles triangle (AB = AC) with angleBAC = 20°.  Point D is on side AC such that angleDBC = 60°.  Point E is on side AB such that angleECB = 50°.  Find, with proof, the measure of angleEDB.


Solution by Construction

Mark K on AC such that angleKBC = 20°.  Draw KB and KE.

Isosceles triangle ABC

angleBEC = angleECB, and so triangleBEC is isosceles with BE = BC.
angleBKC = angleBCK, and so triangleBKC is isosceles with BK = BC.
Therefore BE = BK. angleEBK = 60°, and so triangleEBK is equilateral.
angleBDK = angleDBK = 40° and so triangleBDK is isosceles, with KD = KB = KE.
So triangleKDE is isosceles, with angleEKD = 40°, since angleEKC = 140°.
Therefore angleEDK = 70°, yielding angleEDB = 30°.

Trigonometric Solution

Let angleEDB = x.  Then angleBED = 160° − x, and angleBDC = 40°.

Isosceles triangle ABC

Applying the law of sines (also known as the sine rule) to:
triangleBED,  BE / sin x = BD / sin(160°−x),
triangleBDC,  BC / sin40° = BD / sin80°.

Therefore BD = BE · sin(160°−x) / sin x = BC · sin80° / sin40°

angleBEC = angleECB, and so triangleBEC is isosceles with BE = BC.
Hence sin(160°−x) / sin x = sin80° / sin40°.

Then sin(160°−x) = sin(20°+x), (since sin a = sin(180°−a)),
and sin80° = 2 sin40° cos40°, (since sin2a = 2 sin a cos a.)

Therefore sin(20°+x) = 2 cos40° sin x.
  = sin(x+40°) + sin(x−40°), (since sin a cos b = ½ [sin(a+b) + sin(a−b)].)
Then sin(20°+x) − sin(x−40°) = 2 cos(x−10°) sin30°
  = sin(x+80°), (since sin a = cos(90°−a).)

Hence sin(x+40°) = sin(x+80°).
If x < 180°, the only solution is x + 80° = 180° − (x + 40°), (since sin a = sin(180°−a).)
Hence x = 30°.

Therefore angleEDB = 30°.


Remarks

This deceptively difficult problem dates back to at least 1922, when it appeared in the Mathematical Gazette, Volume 11, p. 173.  It is known as Langley's problem.  See (Adobe) Portable Document Format An Intriguing Geometry Problem for further details.

The problem may be approached using Ceva's Theorem; see the discussion in Trigonometric Form of Ceva's Theorem.


A generalization (4 star)

Consider the case where ABC is an isosceles triangle (AB = AC), with angleBAC = 2a, angleDBC = b, and angleECB = c.  Find angleEDB in terms of a, b, and c.

Answer  -  Solution

Source: Mathematical Gazette; see above.

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