See complete Problem 106

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 106

Subscribe to:
Post Comments (Atom)

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

###
Geometry Problem 106

## Search This Blog

## Blog Archive

## Link List

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

See complete Problem 106

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

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

Construct E on BC such that AE bisects angle BAC, then AE=CE implies triangles ABE and CDE are congruent, thus ABED is cyclic and angle DBC=a, therefore angle ABD=pi-4a.

ReplyDeleteOr

ReplyDeleteextend CA such BA=AE; m(ABE)=m(BEA)=a

BE=BC & DC=AE so AB=BD

x=180-4a

The symmetrical point D about the BC is E, then AB = DC =EC and BD=DE,<BCE=a=<ECB, but <BAC=<ECA=2a and AB=CE then ABEC is isosceles trapezoid (ΒΕ//ΑC ),with <CBE=<ACB=a=<ECB so BE=EC=DC=BD . Then <DBC=<DCB=a, so <BDA=<DBC+<DCB=2a.Therefore x=180-2a-2a=180-4a.

ReplyDeleteIf the perpendicular bisector of BC cuts AC at E, then <AEB=2a, then CE=BE=AB and E=D and x=180-4a

ReplyDelete[For easy typing, I use "a" to replace alpha]

ReplyDelete<BDC=180-x-3a

Consider triangle BDC

sin(180-x-3a)/CD=sina/BD

CD/BD=sin(x+3a)/sina---------(1)

Consider triangle BAD

sin2a/BD=sin(180-x-2a)/AB

AB/BD=sin(x+2a)/sin2a-----------(2)

Since AB=CD, by equating (1) & (2)

sin(x+3a)/sina=sin(x+2a)/sin2a

2cosa=sin(x+2a)/sin(x+3a)

2sin(x+3a)cosa=sin(x+2a)

sin(x+4a)+sin(x+2a)=sin(x+2a)

sin(x+4a)=0

x+4a=0 (rej) or 180

x=180-4a