Proposed Problem

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

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Complete Problem 439

Level: High School, SAT Prep, College geometry

## Sunday, April 25, 2010

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

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angle(ICD) = angle(ICB) = alpha (IC bisector)

ReplyDeleteangle(BAC) = angle(BCA) = x + alpha (Tri.ABC isosceles)

Tri.ADC: x+alpha + 94º + x-alpha = 180º

x = 43º

MIGUE.

name ang DCI = a, DCA = b

ReplyDelete=>

x = a + b , A = 2a + b, CIT = 90 - a ( T tg point )

in ADIC

(a + b) + (90 - a + 47) + ( 43 + 94) + (2a + b) = 360

2a + 2b = 86

a + b = 43

x = 43

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Let BC,CD,DB touch the circle at D',B',C' respectively

ReplyDeletemDIC=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

ang. BAC = ang.BCA

ReplyDeleteang. 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

AB=BC which is given so that angleBAC = angleACB.

ReplyDeletewhich 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.

angleADI=angleADC+angleCDI=94+(1/2)*angleBDC because ID bisects angleBDC as I is the incentre.

which implies that angleADI=94+43=137

but angleADI+angleACI=180

which implies that 137+angleACI=180

so that angleACI=43.

Let < ICD = < ICB = m.

ReplyDeleteThen in Tr. ACD, C + (C - 2m) + 94 = 180 from which we have C - m = 43 = x.

Sumith Peiris

Moratuwa

Sri Lanka