## 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