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.

## Saturday, May 21, 2016

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

Labels:
angle bisector,
circle,
circumcircle,
cyclic quadrilateral,
excenter,
incenter,
tangent,
triangle

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http://s32.postimg.org/679nin5x1/pro_1217.png

ReplyDeleteDenote (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

∠IBL=∠IBC+∠CBL=1/2∠ABC+1/2(180-∠ABC)=1/2×180=90

ReplyDeleteSimilarly，∠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