Concyclic Points in a Triangle
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Let AH_{a} be the A-altitude of ΔABC, AL_{a} the angle bisector of the angle at A, A' the midpoint of BC, P and Q the feet of perpendiculars from B and C on AL_{a}. Prove that the four points P, Q, A' and H_{a} are concyclic.
References
- D. Grinberg, From Baltic Way to Feuerbach - a geometrical excursion, Mathematical Reflections 2, 2006
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Solution
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We are going to establish the required fact by comparing two angles, PA'H_{a} and PQH_{a}. Sometimes,depending on the configuration, the two angles are congruent and sometimes they are supplementary. In both cases the four points P, A', Q, and H_{a} are concyclic, but the derivations are slightly different. Luckily, there is a way to combine the two derivation into one. The device we are going to use is working directed angles modulo 180°.
Since AP is the bisector of the angle at A, P is the midpoint of segment BR with R on AC. In ΔBCR, A'P is a midline and hence parallel to the base: A'P||AC. Depending on whether A' is to the right or left of L_{a} angles ACB and PA'B are either congruent or supplementary:
(CA, CB) = (A'P, A'H_{a}). |
On the other hand, since angles AH_{a}C and AQC are both right, both H_{a} and Q lie on the circle with diameter AC making the four points, A, H_{a}, Q, C, concyclic. It follows that
(CA, CH_{a}) = (QA, QH_{a}). |
But (CA, CB) and (CA, CH_{a}) are one and the same angle and the same holds for (QA, QH_{a}) and (QP, QH_{a}). Therefore,
(A'P, A'H_{a}) = (QP, QH_{a}). |
And the points P, H_{a}, Q, A' are indeed concyclic.
The configuration has other notable properties:
- A'Q||AB.
This is shown exactly as A'P||AC.
- A'P = A'Q.
Indeed, A'P||AC making (PA', PQ) = (AC, AQ). Similarly, from A'Q||AB,
(QA', Q A) = (AB, AQ). But APQ is the bisector of angle at A so thatBAQ = CAQ. It follows that ΔPA'Q is isosceles, so that indeedA'P = A'Q. - ΔQPH_{a} is inversely similar to ΔABC.
Indeed, PQH_{a} = AQH_{a} = ACH_{a} = ACB. Similarly,
QPH_{a} = ABC.
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Copyright © 1996-2018 Alexander Bogomolny