## Thursday, October 29, 2015

### Geometry Problem 1160: Triangle, Three Squares, Area, Equivalent Triangles

Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below for more details.

1. http://s16.postimg.org/u34puqelh/pro_1160.png

Denote S(XYZ)= area of XYZ
Draw altitudes AR and CL and lines per sketch
Observe that triangle B1HC congrurent to CLA ( case ASA) => B1H=CL
We have S(B1BC)= ½ .(BC1 x B1K).... (1)
But B1K= B1H+HK= CL+ BL
Replace this in (1) we have S(B1BC)=1/2. BC1 x(CL+BL)= S(ABC)+1/2. S(BLMC1)
Simillarly S(B2BA2)= S(ABC)+ ½. S(BRSA2)
Note that quadrilateral ACRL is cyclic => BL. BA=BR.BC => S(BLMC1)= S(BRSA2)
Replace it in above expressions we get S(B1BC)= S(B2BA2)

1. I wonder where point H is...

2. H is the projection of B1 over CL extended . See sketch for detail

2. I doubt I would ever come up with such a brilliant proof like Peter's. I proved this with a somewhat dumber approach using algebra.
Area(BB1C1) = BB1*B1C1*sin(∠BB1C1)/2
Area(BB2A2) = BB2*A2B2*sin(∠BB2A2)/2
My approach involves these two steps:
(1) BB1*B1C1 == BB2*A2B2
(2) ∠BB1C1 == ∠BB2A2
They are not trivial but not too hard to reach either.