See complete Problem 107

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

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Geometry Problem 107

Subscribe to:
Post Comments (Atom)

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

###
Geometry Problem 107

## Link List

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

See complete Problem 107

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

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

Construct E on AC such that angles ABC=BEC, then angles CBD=EBD and we have DE/BE=DC/CB=AB/BC so triangles ABC~DEB giving triangle BAD isosceles thus x=pi-4a.

ReplyDeletehttp://geometri-problemleri.blogspot.com/2009/05/problem-24-ve-cozumu_14.html

ReplyDeleteSolution to problem 107.

ReplyDeleteTake E on BC, such that AE is bisector of ang(BAC). Notice that ang(BAE) = ang(DBE) = alpha and both subtend the same segment DE. So ABED is cyclic, ang(BDE) = alpha, tr. BDE is isosceles and BE = DE.

In triangle ABD, ang(BDC) = x + 2.alpha, so

ang(CDE) = x + alpha.

As AB = CD, BE = DE and

ang(ABE) = ang(CDE) = x + alpha,

triangles ABE and CDE are congruent, with

ang(BAE) = ang(DCE) = alpha.

Thus, in triangle ABC,

(x + alpha) + 2.alpha + alpha = 180

and finally x = 180 – 4.alpha.

Let AE be the bisector of < BAC where E is on BC

ReplyDeleteTr. s ABE and CDE are congruent SAS so < EDC = < ABE implying that ABED is cyclic and < DBC = @

So x = 180-4@

Sumith Peiris

Moratuwa

Sri. Lanka

Let E be on BC such that AE bisects < BAC

ReplyDeleteThen ABED is cyclic and so Tr.s ABE and DEC are congruent SAS

Hence AE = EC and so < ACE = @

So x = 180-4@

Sumith Peiris

Moratuwa

Sri Lanka

Bring AE=//DC (AB=DC) then AECD is parallelogram so <BEA=<BAE=<CAE=a then AE is bisector angle BAC.Is <ECB=<DBC=a=<BAE so BECA is isosceles trapezoid or concyclic.

ReplyDeleteTherefore <BCA=<BAE=a and <BDA=2a,so x=180-4a.