See complete Problem 126

Incenter of Triangle, Angle Bisector, Proportions. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, June 23, 2008

### Elearn Geometry Problem 126

Labels:
angle bisector,
area,
curvilinear triangle,
incenter,
incircle,
proportions

Subscribe to:
Post Comments (Atom)

in-radius=r

ReplyDeletee/d=[ABCI]/[AIC]=(ar/2+cr/2)/(br/2)=(a+c)/b

http://img24.imageshack.us/img24/4625/problem126.png

ReplyDeleteFrom B draw a line // to AC

Extend AI and CI to E and F ( see attached sketch)

Note that ∆FBC and ∆ABE are isosceles

So FB=BC=a and AB=BE=c

∆ AIC similar to ∆ EIF ….( Case AA)

So FE/AC=IE/IA=IB/ID

And e/d= (a+c)/b

Solution to problem 126.

ReplyDeleteIn triangle ABC, as BD is the angle bisector of B, then

AB/AD = BC/DC = (AB+BC)/(AD+DC) = (AB+BC)/AC.

In triangle ABD, since AI is the angle bisector of A, then AB/AD = BI/ID.

Hence BI/ID = (AB+BC)/AC, or e/d = (a+c)/b.

To Antonio:

ReplyDeleteIn the enunciate of problem 126, "the angle bisector of A" should be "the angle bisector of B". Am I right?

Thanks, enunciate has been updated.

DeleteConnect the points I and C , then using the angle bisector theorem we get :

ReplyDeletea/e=DC/d

(1.) ad/e=DC

Connect the points I and A , then using the angle bisector theorem we get :

c/e=AD/d

(2.) cd/e=AD

Adding equation 1 and 2 with AD+DC=b :

(ad+cd)/e=AD+DC

(ad+cd)/e=b

(a+c)/b=e/d