## Sunday, April 25, 2010

### Problem 439: Isosceles triangle, Cevian, Incenter, Angles, Circle

Proposed Problem
Click the figure below to see the complete problem 439 about Isosceles triangle, Cevian, Incenter, Angles, Circle.

Complete Problem 439

Level: High School, SAT Prep, College geometry

1. angle(ICD) = angle(ICB) = alpha (IC bisector)
angle(BAC) = angle(BCA) = x + alpha (Tri.ABC isosceles)
Tri.ADC: x+alpha + 94º + x-alpha = 180º
x = 43º

MIGUE.

2. name ang DCI = a, DCA = b
=>
x = a + b , A = 2a + b, CIT = 90 - a ( T tg point )
(a + b) + (90 - a + 47) + ( 43 + 94) + (2a + b) = 360
2a + 2b = 86
a + b = 43

x = 43
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3. Let BC,CD,DB touch the circle at D',B',C' respectively
mDIC=mDIB'+mB'IC=1/2(mC'IB'+mB'ID')=m(IC',BI) <=> m(BI,IC)=mC'ID, with mDC'I=m(AC,IB)=pi/2 => mCDI=mIDB=x
2x+94=180 <=> x=43

4. ang. BAC = ang.BCA
ang. BCA=2x+y =BAC if ICD =x and DCA=y
in tri.ADC ang A +ang C=2x+2y=180-94=86
hence x+y=43

5. AB=BC which is given so that angleBAC = angleACB.
which implies that angleABC=180-2*angleBAC
join ID then in triangle BDC, I is the incentre.
so that angleDIC=90+(1/2)*angleABC but we have angleABC = 180-2*angleBAC which gives angleDIC=180-angleBAC.
but from the above we have, angleDAC+angleDIC=180 so that A,D,I,C are concyclic.
now angleADI+angleACI=180 by angle sum property.
angleADC=94 which is given so that angleBDC=180-94=86 as these angles make a linear pair.