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Geometry Difficulty 4.7 AIME Prove it Italy

Problem:

Let ABCABC be an acute triangle, and let D,ED, E be the feet of the altitudes from A,BA, B. Let A′A' be the midpoint of ADAD, B′B' the midpoint of BEBE. CA′CA' intersects BEBE at XX, CB′CB' intersects ADAD at YY. Prove that there exists a circle passing through the points A′,B′,X,YA', B', X, Y.

Solution

Solution:

The triangles ADCADC and BECBEC are similar, being right triangles with the same angle at CC: it follows that the triangles BB′CBB'C and AA′CAA'C are also similar, and, in particular, that BB′C^=CA′A^\widehat{BB'C} = \widehat{CA'A} holds. There are now two cases: either the quadrilateral A′XB′YA'XB'Y is crossed, or it is not.

Figure 1

In the first case, A′A' and B′B' see the segment XYXY under the same angle.

Figure 2

In the second case, the quadrilateral A′XB′YA'XB'Y has two opposite angles that are supplementary. In both cases, what has been proved is sufficient to establish the cyclicity of the quadrilateral A′XB′YA'XB'Y, that is, that the vertices A′,X,B′,YA', X, B', Y belong to one and the same circle.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.