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Solution to puzzle 123: Right angle and median

Skip restatement of puzzle.Let ABC be a triangle, with AB not equal to AC.  Drop a perpendicular from A to BC, meeting at O.  Let AD be the median joining A to BC.  If angleOAB = angleCAD, show that angleCAB is a right angle.

Let AB < AC.  Let E be the midpoint of AC.  Draw OE and DE.
Let angleOAB = angleCAD = x, and let angleDAO = y.

Triangle ABC, with perpendicular AO and median AD, as described above. Draw OE and DE, where E is the midpoint of AC. Angle OAB = angle CAD = x, and angle DAO = y.

Since E is the midpoint of AC, a line from E, parallel to BC, will bisect line segment AO.
Hence OE = AE, and so angleAOE = angleEAO = x + y.
(Alternatively, consider the semicircle with diameter AC, passing through O.)

Since D and E are midpoints, DE is parallel to BA, and so, considering alternate interior angles, angleADE = angleBAD = x + y.
That is, angleAOE = angleADE.
We deduce that points A, O, D, and E are concyclic; that is, they lie on a circle.

(This follows from the result that the locus of all points from which a given line segment subtends equal angles is a circle.  See Munching on Inscribed Angles; reference 1, below.  We will use a converse of this result below: all angles inscribed in a circle, subtended by the same chord and on the same side of the chord, are equal.)

Triangle ABC, with perpendicular AO and median AD, as described above. Draw OE and DE, where E is the midpoint of AC. Angle OAB = angle CAD = x, and angle DAO = y. Angle AOE = angle ADE = x+y. Hence draw circle touching A, O, D, and E.

Now consider chord DE.
The angle subtended at O equals the angle subtended at A.
That is, angleEOD = angleEAD = x.
Hence pi/2 = angleAOD = angleAOE + angleEOD = 2x + y = angleCAB.

Therefore, angleCAB is a right angle, which was to be proved.

Remarks

The converse of this result, that if angleCAB is a right angle then angleOAB = angleCAD, is also true; see Right Triangle Equal Angles on mathschallenge.net.

Isosceles triangle ABC, with AB = AC, and hence O = D, so that perpendicular AO and median AD coincide.

It may be wondered why we needed the condition AB not equal to AC for this result to hold.  Well, if AB = AC, we have an isosceles triangle, in which points O and D coincide, and in which angleCAB can take any value in the open interval (0, pi).  The above proof breaks down when we try to consider triangleODA!

However, the proof holds if O and D are arbitrarily close, yet distinct.  You may wish to try drawing the diagram for the case AB = 20, AC = 21, BC = 29.

In the proof above, having shown that A, O, D, and E are concyclic, we could have noted that opposite angles of a cyclic quadrilateral sum to pi, and hence angleDEA = pi/2.  Then, since DE is parallel to BA, we immediately obtain angleCAB = pi/2.

Alternatively, we could have noted that angleAOD = pi/2 implies DA is a diameter implies angleDEA = pi/2 implies angleCAB = pi/2.

Alternative proof

Michael Hemy sent the following solution, which does not make use of an auxiliary circle.

Let AB < AC.  Let angleOAB = angleCAD = x, and let angleDCA = y.
Then angleABO = 90° − x, angleODA = x + y, and angleDAO = 90° − x − y.

Triangle ABC, with perpendicular AO and median AD, as described above. Draw OE and DE, where E is the midpoint of AC. Angle OAB = angle CAD = x, and angle DCA = y.

Applying the law of sines (also known as the sine rule):
triangleABD,  AD / sin(90° − x) = BD / sin(90° − y)
triangleADC,  AD / sin(y) = DC / sin(x) = BD / sin(x), since BD = DC.

Hence sin(90° − x) / sin(y) = sin(90° − y) / sin(x).
Since sin(90° − a) = cos(a), we have sin(x) cos(x) = sin(y) cos(y), and so ½ sin(2x) = ½ sin(2y).
Therefore 2x = 2y or 2x + 2y = 180°.
The latter is impossible, as angleADC > 90°.  Hence x = y.

Therefore, angleCAB is a right angle, which was to be proved.

Further reading

  1. Munching on Inscribed Angles
  2. Munching on Circles
  3. [Java] Inscribed and Central Angles in a Circle
  4. Ptolemy's Theorem
  5. Ptolemy's Theorem and Interpolation

Source: Arunabha Biswas

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