Geometry : 103

123. Right angle and median 

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.

Solution to puzzle 123: Right angle and median

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 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.

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