Chapter 1: Problem 2
Cevians perpendicular to the opposite sides are concurrent.
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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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Any triangle having two equal medians is isosceles.
If a cevian \(A Q\) of an equilateral triangle \(A B C\) is extended to meet the circumcircle at \(P\), then $$ \frac{1}{P B}+\frac{1}{P C}=\frac{1}{P Q} $$
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 \(A X\) be a cevian of length \(p\), dividing \(B C\) into segments \(B X=m\) and \(X C=\hbar\), as in Figure \(1.2 \mathrm{C}\). Then $$ a\left(p^{2}+m n\right)=b^{2} m+c^{2} n $$ Hint: Add expressions for the cosines of the two supplementary angles at \(X\) in terms of the sides of \(\triangle A B X\) and \(\triangle C A X\), respectively. This result is called Stewart's theorem, after M. Stewart, who stated it in 1746 . It was probably discovered by Archimedes about 300 B.C., but the first known proof is by R. Simson, 1751 .
The circumcircle of \(\triangle A B C\) is the nine-point circle of \(\triangle I_{a} I_{s} I_{e}\).
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