## Monday, May 19, 2008

### Geometry Problem 104

See complete Problem 104
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

1. Let I be an incenter fo the trinagle ABD. Then
angle 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
x= angle of AEB=90+beta-alpha.

2009.9.20
From seoul, South Korea.
Bae deok rak
bdr@korea.com

2. 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.
Let 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.

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

sumith peiris
Moratuwa
Sri Lanka

1. How CA will be internal bisector pls prove

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

Similarly 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