Chapter 1: Problem 2
Cevians perpendicular to the opposite sides are concurrent.
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These are the key concepts you need to understand to accurately answer the question.
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Chapter 1: Problem 2
Cevians perpendicular to the opposite sides are concurrent.
These are the key concepts you need to understand to accurately answer the question.
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If a cevian \(A Q\) of an equilateral triangle \(A B C\) is extended to meet the circumcircle at \(P\), then $$ \frac{1}{P B}+\frac{1}{P C}=\frac{1}{P Q} $$
The product of two sides of a triangle is equal to the product of the circumdiameter and the altitude on the third side.
The orthocenter of an obtuse-angled triangle is an excenter of its orthic triangle.
Let \(A B C\) and \(A^{\prime} B^{\prime} C^{\prime}\) be two non-congruent triangles whose sides are respectively parallel, as in Figure 1.2B. Then the three lines \(A A^{\prime}, B B^{\prime}\), \(C C^{\prime}\) (extended) are concurrent. (Such triangles are said to be homothetic. We shall consider them further in Section 4.7.)
Let three congruent circles with one common point meet again in three points \(A, B, C\). Then the common radius of the three given circles is equal to the circumradius of \(\triangle A B C\), and their common point is its orthocenter.
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