Proposed Problem

Click the figure below to see the complete problem 440 about Triangle, Incircle, Incenter, Angle Bisector, Tangency Points, Circle, Angles.

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Complete Problem 440

Level: High School, SAT Prep, College geometry

## Wednesday, April 28, 2010

### Problem 440: Triangle, Incircle, Incenter, Angle Bisector, Tangency Points, Circle, Angles

Labels:
angle,
angle bisector,
circle,
incenter,
incircle,
tangency point,
triangle

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join I to E and to F

ReplyDeleteFIE = 180°- 44 = 136° => IFE = IEF = 44:2 = 22°

=> AFE = 90° + 22° = 112°

A/2 + B/2 = 136:2 =68°

=>

AFE + AIG = 112°+ 68° = 180° => AIGF cyclic =>

x = IFE ( have same arc IG )

x = 22°

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mBIA=mEIF/2=mEFF=mEFA

ReplyDeleteAIGF, IECF concyclic

x=mGFI=mECI=44/2=22

A + B = 136 and so A/2 + B/2 = 68.

ReplyDeleteNow IECF being cyclic < IEF = 22 and considering the angles of Tr. BEG, < BGE = 180 - B/2 - 90 - 22 = A/2. Hence AIGF is cyclic and so x = < IFG = 22

Sumith Peiris

Moratuwa

Sri Lanka

Even if the value of C is not given we can similarly show that AG is perpendicular to BG

ReplyDelete