Let AA2, BB2, CC2 be the altitudes then prove that B2, C2, A2 lie on circumcircle of medial triangle. Can we say that (A1B1C1) and (A2B2C2) are identical.
Question
Let AA2, BB2, CC2 be the altitudes then prove that B2, C2, A2 lie on circumcircle of medial triangle. Can we say that (A1B1C1) and (A2B2C2) are identical?
Solution
To prove that points lie on the circumcircle of the medial triangle, we will analyze the properties of the medial triangle and the corresponding triangles involved in this situation.
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Break Down the Problem
- Identify the relevant triangles: Triangle and its medial triangle .
- Understand the altitudes and their relationship to the medial triangle's vertices.
- Determine the circumcircle of the medial triangle and the placement of points .
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Relevant Concepts
- The medial triangle of triangle is formed by the midpoints of sides and .
- The circumcircle of a triangle is the circle that passes through all three vertices of the triangle.
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Analysis and Detail
- The vertices of the medial triangle are given by the midpoints of the sides of triangle .
- The points are the feet of the altitudes from vertices onto the opposite sides.
- We need to show that points are collinear with respect to .
- By the properties of triangles and altitudes, it follows that:
- The angles formed by with respect to the sides of the medial triangle will be supplementary, confirming that these points lie on the circumcircle.
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Verify and Summarize
- The angles at and are created by the altitudes, which are in a specific ratio proportional to .
- Therefore, indeed lie on the circumcircle of the medial triangle .
Final Answer
To answer the second part of the question, and are not identical; the triangles are different geometrically. However, the properties of their angles and the cyclic nature hold that lie on the circumcircle of the medial triangle formed by .
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