## Tuesday, October 26, 2010

### Problem 532: Triangle, Circumcircle, Equal Angles, Perpendicular, Area

Geometry Problem
Click the figure below to see the complete problem 532 about Triangle, Circumcircle, Equal Angles, Perpendicular, Area.

Complete Problem 532

Level: High School, SAT Prep, College geometry

1. Let /_DBE=β. Then Tr. ABC=AB*BC*sin(2α+β)/2 or Tr. ABC=(1/2)(AB*BC*sin(α+β)*cos(α)+AB*BC*cos(α+β)sin(α). By sine rule,BC/BD=sin(A+α)/sinC. Also BE/BD=sin(A+α)/sinC. In other words, BC/BD=BE/AB or AB*BC=BD*BE which makes Tr ABC==(1/2)(BD*BE*sin(α+β)*cos(β)+BD*BE*cos(α+β)sin(α). But cos(α)=BF/BD & cos(α+β)=BG/BD which when substituted in our eqn. for Tr. ABC gives us:Tr.ABC=BE.BF*sin(α+β)/2+BE*BG*sin(α)/2 = Tr. BFE+Tr.GBE. In other words, Tr.ABC=Quad.BFEG which means, A1+Area BFMNG=A2+A3+Area BFMNG or A1 = A2 + A3
Ajit: ajitathle@gmail.com

2. 1. Let BD=d
Let H and H’ are the projection of G and F over BE
We have BH=d.cos(α).cos(DBG) and BH’=d.cos(α).cos(FBE)
But angle (DBG)= angle( FBE) so BH=BH’ and BE perpendicular to FG

2. Triangle ABD similar to triangle BEC (case AA)
And AB/BE=BD/BC so AB.AC=BD.BE (1)

3. Quadrilateral DFBG is cyclic and BD is a diameter
BD=FG/sin(ABC)

4. Replace BD=FG/sin(ABC) in (1) we have AB.AC.sin(ABC)= FG.BE
That means that Area(ABC)=Area(FBGE) and we will have A1=A2+A3

Peter Tran

3. Let H be the projection of B over AC
Let angle DBE=β

The quadrilateral BDHG is inscribed in a circle of diameter BD
Therefore angle GHC=angle DBG=α+β

Because quadrilateral ABCE is inscribed in a circle C(1), angle ABE = angle ACE =α+β

So HG is parallel to EC

A quadrilateral FBHD is inscribed in a circle, therefore angle FHD = angle FBD =α

Angle CAE = angle CBE =α

Then FH is parallel to AE

That means that
Area FEH = Area FAH
Area HGE = Area HGC

Then Area of the quadrilateral BFEG = Area of the triangle ABC

we will have A1=A2+A3

As our mentor says “It is possible to provide a proof using only elementary geometry!”