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How to Prove It A S How to E Prove It P In this episode of “How To Prove It”, we prove a simple yet surprising collinearity associated with any D B C ClassRoom triangle. We do so in two different ways and then contrast the two proofs. It would be instructive for a Q student to study these proofs and also the remarks made at the end. F R H Shailesh A Shirali Figure 2 In ABC, let the feet of the altitudes from A, B, C be D, E, F △ In case you are wondering how the configuration supplementary. We show this as follows. Since respectively. Select any one altitude, say AD, and from its foot will change if the triangle is obtuse, please study quadrilateral PBDQ is cyclic, it follows that (D), drop perpendiculars to the other two sides (AB, AC) and to Figure 2 (in which B is obtuse). Many of the PQD = 180◦ B. Since quadrilateral QDRH the other two altitudes (BE, CF). Let the feet of these ̸ ̸ − ̸ points now fall outside the given triangle. The is cyclic, it follows that RQD = RHD. But perpendiculars be P, S and Q, R, as shown in Figure 1. Then the ̸ ̸ conclusion however stays intact. RHD = FBD (this follows because points P, Q, R, S lie in a straight line. ̸ ̸ quadrilateral FBDH is cyclic), i.e., RHD = B. We offer two proofs of the proposition. The reader ̸ ̸ It follows that PQD and RQD are A is invited to judge the relative merits of each proof. ̸ ̸ supplementary. Hence points P, Q, R lie in a AD BC straight line. ⊥ First proof BE CA In exactly the same way we show that points ⊥ This proof uses nothing more than ‘angle chasing’. CF AB Q, R, S lie in a straight line. It follows that all four E ⊥ We start by redrawing the figure, taking care to DP AB points P, Q, R, S lie in a straight line, as ⊥ use different colours for segments PQ, QR and RS DQ BE required. F S ⊥ (see Figure 3). The purpose of doing this is simply H DR CF to avoid the error of implicitly assuming the very R ⊥ DS AC thing you are trying to prove—an error that is Second proof Q ⊥ P H: orthocentre particularly easy to make in geometry. This time we shall use coordinates. Adopt a system of coordinates in which line BC represents the Note the presence of a number of cyclic x-axis, and line AD represents the y-axis; so D quadrilaterals; this is natural, given the number of B D C serves as the origin. Let the coordinates of the perpendiculars that have been drawn. Note also Figure 1 vertices be as follows: A =(0, a), B =(b, 0), that we have coloured some of the quadrilaterals C =(c, 0). Note that a, b, c do not represent the differently. Keywords: Altitudes, collinearity, cyclic quadrilateral, angle lengths of the sides of the triangle. We now chasing, supplementary, coordinate, slope, linear equation, pure To prove that points P, Q, R lie in a straight line, it compute the coordinates of the points P, Q, R, S geometry, proof suffices to show that ̸PQD and ̸RQD are in terms of a, b, c. 5859 At Right Angles || Vol. 5, No. 2, July 2016 Vol. 5, No. 2, July 2016 || At Right Angles 5859 1 A S E P D B C Q F R H Figure 2 In case you are wondering how the configuration supplementary. We show this as follows. Since will change if the triangle is obtuse, please study quadrilateral PBDQ is cyclic, it follows that Figure 2 (in which B is obtuse). Many of the PQD = 180◦ B. Since quadrilateral QDRH ̸ ̸ − ̸ points now fall outside the given triangle. The is cyclic, it follows that ̸RQD = ̸RHD. But conclusion however stays intact. ̸RHD = ̸FBD (this follows because quadrilateral FBDH is cyclic), i.e., RHD = B. We offer two proofs of the proposition. The reader ̸ ̸ It follows that PQD and RQD are is invited to judge the relative merits of each proof. ̸ ̸ supplementary. Hence points P, Q, R lie in a straight line. First proof In exactly the same way we show that points This proof uses nothing more than ‘angle chasing’. Q, R, S lie in a straight line. It follows that all four We start by redrawing the figure, taking care to points P, Q, R, S lie in a straight line, as use different colours for segments PQ, QR and RS required. (see Figure 3). The purpose of doing this is simply to avoid the error of implicitly assuming the very thing you are trying to prove—an error that is Second proof particularly easy to make in geometry. This time we shall use coordinates. Adopt a system of coordinates in which line BC represents the Note the presence of a number of cyclic x-axis, and line AD represents the y-axis; so D quadrilaterals; this is natural, given the number of serves as the origin. Let the coordinates of the perpendiculars that have been drawn. Note also vertices be as follows: A =(0, a), B =(b, 0), that we have coloured some of the quadrilaterals C =(c, 0). Note that a, b, c do not represent the differently. lengths of the sides of the triangle. We now To prove that points P, Q, R lie in a straight line, it compute the coordinates of the points P, Q, R, S suffices to show that ̸PQD and ̸RQD are in terms of a, b, c. 59 At Right Angles | Vol. 5, No. 2, July 2016 Vol. 5, No. 2, July 2016 | At Right Angles 59 A This particular problem viewed as a ‘pure involved. For example, it is not true to say that no geometry problem’ is not too difficult; it yields to intuitive feel is involved in the application of elementary angle chasing. However, let us coordinates. A very important first step when one consider the matter more generally. uses coordinates is the choice of axes; they must be chosen in such a way as to minimise the number Problems in geometry can sometimes be very of symbols being used, and exploit to the challenging, because the figure typically does not maximum the symmetries implicit in the figure. E give any clue or hint as to the direction in which In the above proof, note how D was made the one must proceed. In general, the difficulty is that origin of the coordinate system, with BC and AD one is not able to ‘see’ the key elements contained F as the axes; this has clearly been done keeping in S in the figure. The solution may depend on seeing mind the number of lines of the figure which pass H that a particular quadrilateral is cyclic; but the R through D. Next, note the symbols used for the quadrilateral may be well-concealed within the Q coordinates of B and C, namely: (b, 0) and (c, 0). figure, its sides not standing out in any way. Or The beginner, noting that in the figure B lies to the P the solution may depend on seeing that two ‘left’ or negative side of D while C lies to the ‘right’ particular angles are equal; but the arms of the or positive side of D, may be tempted to write: angles may not stand out in any way. For such B =( b, 0) and C =(c, 0). But this would be B D C reasons, it helps if we mark the figure suitably, in − Figure 3 quite unnecessary, because b and c can be either advance: systematically look for pairs of angles positive or negative; no signs have been fixed as which are equal to one another, and mark them yet. Also, such a use of symbols would have The slopes of the lines AB and AC are a/b and Now, by a routine calculation (but we omit the so; systematically look for pairs of segments that − spoiled the symmetry which we see at present. a/c, hence the slopes of CF and DP are b/a, and details), we find that the slopes of the segments are congruent to one another, and mark them so; − Writing B =( b, 0) would have been contrary to the slopes of BE and DS are c/a. It follows that PQ, QR and RS are all equal to the following similarly, mark pairs of lines which are parallel to − the very spirit of algebra. During the course of the the equations of the various lines in the figure are expression: one another, or perpendicular to one another, and solution we remarked: “The coordinates of S may as follows: a(b + c) mark them so; and so on. The judicious use of . be obtained from those of P by switching the roles a2 bc colour can help in carrying out these markings. − of b and c.” But this would not have been possible Equation of AB: ax + by = ab Hence the points P, Q, R, S lie in a straight Obviously, all these steps by themselves will not • had we written B =( b, 0). Equation of AC: ax + cy = ac line. guarantee anything; but they can ease the path for − • us. Often they do, so it is worth taking these steps. Another noteworthy point is the following. A Equation of BE: cx ay = bc • − Remarks. The following remarks may be of proof based on purely geometrical considerations The use of coordinates to solve problems in Equation of CF: bx ay = bc interest and should be taken note of: typically starts by drawing a diagram and marking • − geometry is generally not recommended = various relationships on it.
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