## Saturday, May 21, 2016

### Geometry Problem 1217: Triangle, Circle, Excenter, Incenter, Angle Bisector, Cyclic Quadrilateral, Circumcircle, Tangent Line

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1217. 1. http://s32.postimg.org/679nin5x1/pro_1217.png
Denote (XYZ) angle (XYZ)
We have (IBL)=(ICL)= 90 => quạ BICL is cyclic
We have (BIK)= 1/2A + 1/2B
(BJK)=1/2(BEC)+1/4B= ½(A+B/2) + 1/4B= 1/2A+1/2B = (BIK) => qua (BIJK) is cyclic
Since quạ BIJK cyclic => (IBJ)=(JBQ)=(IKJ) => quạ BPQK is cyclic
(RAL)=(RBL)= 90 => qua. ABLR is cyclic

Since quạ. BICL is cyclic= > (JLB)= (BCI)= C/2
Let x= (IBP)=(BPQ)=(PKQ)=(MKL)
We have (BKI)=(BJI)= x+C/2
In triangle KML we have (BMK)= x+C/2
Triangles BKl similar to trị KML => LK ^2=LM.LB
So LK tangent to circumcircle of triangle BKM

2. ∠IBL=∠IBC+∠CBL=1/2∠ABC+1/2(180-∠ABC)=1/2×180=90
Similarly，∠ICL=90
∠IBL+∠ICL=90+90=180
BICL concyclic
∠IKJ=∠AKE=∠KEC-∠KAC=1/2∠BEC—1/2∠BAC
=1/2∠ABE=1/2∠EBC=∠IBJ
BIJK concyclic
∠PBQ=∠IBJ=∠IKJ=∠PKQ
BPQK concyclic
BICL concyclic,similarly AICK concyclic
∠BLC=∠RIC=∠RAC
∠BLR+∠BAR=∠BLC+∠BAR=∠RAC+∠BAR=∠RAC+∠BAC+∠CAR=180
ABLRconcyclic
∠BMK=∠MLK+∠LKM=∠PKQ+∠BCI=∠PBQ+∠BCI=∠BJI=∠BKI
∴AL is tangent to circumcircle of triangle BKM