Let ABC be a triangle, with AB AC. Drop a perpendicular from A to BC, meeting at O. Let AD be the median joining A to BC. If OAB = CAD, show that CAB is a right angle.

Let AB < AC. Let E be the midpoint of AC. Draw OE and DE.

Let OAB = CAD = x, and let 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 AOE = EAO = 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, ADE = BAD = x + y.

That is, AOE = ADE.

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.)

Now consider chord DE.

The angle subtended at O equals the angle subtended at A.

That is, EOD = EAD = x.

Hence /2 = AOD = AOE + EOD = 2x + y = CAB.

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

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

It may be wondered why we needed the condition AB 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 CAB can take any value in the open interval (0, ). The above proof breaks down when we try to consider ODA! 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 , and hence DEA = /2. Then, since DE is parallel to BA, we immediately obtain CAB = /2.

Alternatively, we could have noted that AOD = /2 DA is a diameter DEA = /2 CAB = /2.

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

Let AB < AC. Let OAB = CAD = x, and let DCA = y.

Then ABO = 90° − x, ODA = x + y, and DAO = 90° − x − y.

Applying the law of sines (also known as the sine rule):

ABD, AD / sin(90° − x) = BD / sin(90° − y)

ADC, 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 ADC > 90°. Hence x = y.

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

- Munching on Inscribed Angles
- Munching on Circles
- [Java] Inscribed and Central Angles in a Circle
- Ptolemy's Theorem
- Ptolemy's Theorem and Interpolation

Source: Arunabha Biswas