Browse · MathNet
PrintSilk Road Mathematics Competition
geometry
Problem
In triangle the angle bisectors of and intersect the sides and at the points and , respectively, and the circumcircle of the triangle at the points and , respectively. Let be the point of intersection of and , and be the incenter of triangle . Prove that passes through the midpoint of .
Solution
Let's define notations of the points and angles as in the figure. Then and
By the law of sines in and we easily obtain: Similarly, in we get: Then By Ceva's theorem in we have In and , using (2) we obtain In and , using (3) we get
By the law of sines in and we easily obtain: Similarly, in we get: Then By Ceva's theorem in we have In and , using (2) we obtain In and , using (3) we get
Techniques
Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleTriangle trigonometryCeva's theoremAngle chasing