The incircle of a triangle $ ABC$ is tangent to its sides $ AB,BC,CA$ at $ M,N,K,$ respectively. A line $ l$ through the midpoint $ D$ of $ AC$ is parallel to $ MN$ and intersects the lines $ BC$ and $ BD$ at $ T$ and $ S$, respectively. Prove that $ TC=KD=AS.$
Problem
Source: Ukraine 1997 grade 9
Tags: geometry, geometry proposed
plane geometry
18.07.2009 20:07
I think S in the intersection of BA and the parallal line 1.it is obvious KD=1/2(a-c) 2.using sin law we can calculate CK=AS then CK+AS=BS-c+a-BT=c-a CK=AS=KD=(a-c)/2
Attachments:
Victory.US
18.07.2009 20:26
i think this problem is not difficult, . we can compute the length of $ MN$ and use the Thales 's theorem. it's Ok
Sashsiam_2
18.07.2009 21:29
Draw a line through A parallel to MN, and let it meet BC at P. Note that BMN is isosceles. We get AS=PT. Since M is midpoint of AC we have PT=TC, so AS=TC as desired. Now note that CN=CK, and we have CN=2TC+NP and CK=2CD-NP (to see the last assertion observe that AK=AM=NP). This yields AD=CD=TC+NP=NT=MS. Subtract AM=AK and the result follows.