Proposed Problem

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

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Complete Geometry Problem 378

Level: High School, SAT Prep, College geometry

## Wednesday, October 28, 2009

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

Labels:
angle bisector,
congruence,
incenter,
isosceles,
parallel,
triangle

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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).

ReplyDeletedraw IQ, IT radius and IP //BD

ReplyDeletetr IDT and IPQ are congrunt ( IQ=IT, angTID=ang PIQ)

=>DBPI is rombus

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

Note That:

ReplyDelete1. 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

Too easy!

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

Sumith Peiris

Moratuwa

Sri Lanka