Chapter 1: Problem 3
The orthocenter of an obtuse-angled triangle is an excenter of its orthic triangle.
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Chapter 1: Problem 3
The orthocenter of an obtuse-angled triangle is an excenter of its orthic triangle.
These are the key concepts you need to understand to accurately answer the question.
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If lines \(P B\) and \(P D\), outside a parallelogram \(A B C D\), make equal angles with the sides \(B C\) and \(D C\), respectively, as in Figure \(1.9 \mathrm{D}\), then \(\angle C P B=\angle D P A\). (Of course, this is a plane figure, not three dimensionall)
The product of two sides of a triangle is equal to the product of the circumdiameter and the altitude on the third side.
The circumcenter and orthocenter of an obtuse-angled triangle lie outside the 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.)
Cevians perpendicular to the opposite sides are concurrent.
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