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See complete Problem 104Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Let I be an incenter fo the trinagle ABD. Thenangle 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 hencethe 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 sox= angle of AEB=90+beta-alpha.2009.9.20From seoul, South Korea.Bae deok firstname.lastname@example.org
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 haveang(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.sumith peirisMoratuwaSri Lanka