Welcome to Subhanil Sir's portal: a challenging geometry problem

Let ABC be an isosceles triangle (AB = AC) with

BAC = 20°. Point D is on side AC such that

DBC = 60°. Point E is on side AB such that

ECB = 50°. Find, with proof, the measure of

EDB.

Solution by Construction

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

BEC = ECB, and so BEC is isosceles with BE = BC.
BKC = BCK, and so BKC is isosceles with BK = BC.
Therefore BE = BK. EBK = 60°, and so EBK is equilateral.
BDK = DBK = 40° and so BDK is isosceles, with KD = KB = KE.
So KDE is isosceles, with EKD = 40°, since EKC = 140°.
Therefore EDK = 70°, yielding EDB = 30°.

Trigonometric Solution

Let EDB = x. Then BED = 160° − x, and BDC = 40°.

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

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

BEC = ECB, and so BEC 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 EDB = 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 An Intriguing Geometry Problem for further details.