Chapter 7: Problem 14
Show that $$ |\angle A B C|+|\angle B C A|+|\angle C A B|=\pi $$ for any \(\triangle A B C\).
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Chapter 7: Problem 14
Show that $$ |\angle A B C|+|\angle B C A|+|\angle C A B|=\pi $$ for any \(\triangle A B C\).
These are the key concepts you need to understand to accurately answer the question.
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In any \(\triangle A B C,\) we have $$ \angle A B C+\angle B C A+\angle C A B \equiv \pi $$
Let \(\ell\) be a line, \(Q \in \ell,\) and \(P\) is an arbitrary point. Suppose \(O\) is the midpoint of \([P Q] .\) Then a line \(m\) passing thru \(P\) is parallel to \(\ell\) if and only if \(m\) is a reflection of \(\ell\) across \(\mathrm{O}\).
Let \(\triangle A B C\) be a nondegenerate triangle, and \(P\) lies between \(A\) and \(B\). Suppose that a line \(\ell\) passes thru \(P\) and is parallel to \((A C) .\) Show that \(\ell\) crosses the side \([B C]\) at another point, say \(Q,\) and $$ \triangle A B C \sim \triangle P B Q $$ In particular, $$ \frac{P B}{A B}=\frac{Q B}{C B} $$
Let \(\triangle A B C\) be an isosceles nondegenerate triangle with base \([A C]\). Assume that a circle is passing thru \(A\), centered at a point on \([A B]\), and tangent to \((B C)\) at the point \(X\). Show that \(\angle C A X=\pm \frac{\pi}{4}\).
Any parallelogram is centrally symmetric with respect to a midpoint of one of its diagolals. In particular, if \(\square A B C D\) is a parallelogram, then (a) its diagonals \([A C]\) and \([B D]\) intersect each other at their midpoints; (b) \(\angle A B C=\angle C D A\) (c) \(A B=C D\).
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