See complete Problem 104

Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Geometry Problem 104

Subscribe to:
Post Comments (Atom)

skip to main |
skip to sidebar
## Monday, May 19, 2008

###
Geometry Problem 104

## Search This Blog

## Blog Archive

## Link List

Online Geometry theorems, problems, solutions, and related topics.

See complete Problem 104

Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

Let I be an incenter fo the trinagle ABD. Then

ReplyDeleteangle of IBC = angle of IDC=90 degree and so the 4-points B,I,D,C on a circle, that is,

the quadrilateral IDCB is a cycle.

Hence, angle of BCD = alpha +beta and

angle of 180- alpha -beta= 90+((angle of )/2).

Furthermore,

angle of DIC=90-alpha

=(angle of A)/2 +(angle of IDA) and hence

the points A, I, E, C are on the same line, that is,

the points A, I, E, C are on the extended diagonal of the cicumcircle of the

quadrilateral IDCB. and so

x= angle of AEB=90+beta-alpha.

2009.9.20

From seoul, South Korea.

Bae deok rak

bdr@korea.com

This solution to problem 104 follows the same idea of Bae Deok Rak. I think that I’ve drawn it up in a clearer way.

ReplyDeleteLet I be the incenter of triangle ABD. Then ang(IBC) = ang(IDC) = 90, the quadrilateral BCDI is cyclic, ang(BCI) = beta and ang(DCI) = alpha.

We must prove that lines AC and IC are the same line.

In triangle ADI, ang(AID) = 180 – A/2 – beta, and in triangle DIC, ang(DIC) = 90 – alpha. So, ang(AID) + ang(DIC) =

= 270 – A/2 – alpha – beta (1).

In triangle ABD, A = 180 – 2.alpha – 2.beta, or A/2 = 90 – alpha – beta. Therefore, in (1), we get ang(AID) + ang(DIC) =

= 270 – (90 – alpha – beta) – alpha – beta = 180.

This conclusion shows that A, I, E and C are collinear points.

Finally, in triangle BCE, we have

ang(AEB) = ang(DBC) + ang(BCI) = (90 – alpha) +beta = 90 + beta – alpha.

BC and DC are external bisectors of Tr ABD hence CA is the internal bisector and the result follows.

ReplyDeletesumith peiris

Moratuwa

Sri Lanka

How CA will be internal bisector pls prove

DeleteExtend AB to P, < PBC = 90 - alpha so BC is an external bisector of Tr. ABD

DeleteSimilarly DC is an external bisector

Hence C is an excentre of Tr. ABD

So CA must be an internal bisector

Hope it’s clear now