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Geometry and Trigonometry
Fig. 2.29
Problem 2.59 Let M be a point in the interior of triangle ABC. Three lines are drawn through M, parallel to triangle’s sides, determining three trapezoids. One draws a diagonal in each trapezoid such that they have no common endpoints, dividing thus ABC into seven parts, four of them being triangles (see Fig. 2.29). Prove that the area of one of the four triangles equals the sum of the areas of the other three.
2.7 Quadrilaterals with an Inscribed Circle Everybody knows that in any triangle one can inscribe a circle whose center is at the intersection point of the angles’ bisectors. What can we say about a quadrilateral? If there exists a circle touching all the quadrilateral’s sides, then its center is equidistant from them, hence it lies on all four angle bisectors. We deduce that a necessary and sufficient condition for the existence of an inscribed circle is that the quadrilateral’s angle bisectors are concurrent (in fact, this works for arbitrary convex polygons). This does not happen in every quadrilateral. We can always draw a circle tangent to three of the four sides (its center being the point of intersection of two of the bisectors) (Fig. 2.30). If three of the four angle bisectors meet at one point, it is easy to see that the fourth one will also pass through that point and that a circle can be inscribed in the quadrilateral. Another necessary and sufficient condition for the existence of an inscribed circle in a quadrilateral is given by the following theorem, due to Pithot. Theorem Let ABCD be a convex quadrilateral. There exists a circle inscribed in ABCD if and only if: AB + CD = AD + BC. Proof Suppose there exists a circle inscribed in the quadrilateral, touching the sides AB, BC, CD and DA at the points K, L, M, N , respectively. Then, since the tan-