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.

Solution

References

D. Grinberg, From Baltic Way to Feuerbach - a geometrical excursion, Mathematical Reflections 2, 2006

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_aC 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 that BAQ = CAQ. It follows that ΔPA'Q is isosceles, so that indeed A'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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