Wednesday, October 28, 2009

Problem 378. Triangle, Incenter, Parallel to a side, Angle, Congruence

Proposed Problem
Click the figure below to see the complete problem 378 about Triangle, Incenter, Parallel to a side, Angle, Congruence.

Problem 378. Triangle, Incenter, Parallel to a side, Angle, Congruence.
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Complete Geometry Problem 378
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 10:13 AM
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4 comments:

  1. Valdir Marques Faria, Colégio VisãoOctober 30, 2009 at 11:11 AM

    Como DE//CB, temos que Ang(DIB) = Ang(CBI) e Ang(EIC) = Ang(BCI). Vejam que I é o incentro do triângulo ABC, então CI e BI são bissetrizes dos ângulos ACB e ABC. Assim os triângulos CEI e BEI são isósceles sendo BD = DI e IE = EC. Assim, como DE = DI + IE, então DE = CD + EB ( demonstrado).

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  2. c .t . e. oDecember 19, 2009 at 8:00 AM

    draw IQ, IT radius and IP //BD

    tr IDT and IPQ are congrunt ( IQ=IT, angTID=ang PIQ)
    =>DBPI is rombus

    at the same way IKCE (IK//EC),rombus

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  3. Peter TranJune 18, 2010 at 4:02 PM

    Note That:
    1. Area of Triangle BIE= Area of Triangle CIE ( same base and height)
    2. Area of triangle DBE= area of Tri. BDI + area of Tri. BIE
    3. From step 1 and 2 , Write formula of area of each triangle .we will conclude that DE=DB+CE

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  4. Sumith PeirisAugust 25, 2015 at 4:07 AM

    Too easy!

    Tr.s IDB & IEC are both isoceles and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

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