## Friday, July 15, 2011

### Problem 634: Right Triangle, Altitude, Incenters, Areas

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

Click the figure below to see the complete problem 634.

1. Note that Triangles ABC, BHC and ABH are similar triangles ( Case AA)
In each triangle , ratio of height from 90 degrees angle/ radius of incenter circle = h1/r1= h2/r2=h3/r3
But we have S(ABC)=S(ABH)+S(BHC)
So S(AEB)+S(BFC)=S(ABC)

Peter Tran

2. Call α = angle A, AB = c, BC = a and AC = b.
Because AE, BF and CD are bisectors, it is easy to see that ∆AEB, ∆BFC and ∆ACD all have angles ½α, 45º-½α and 135º, so they are all similar triangles.
factor ∆ACD → ∆ABE = c/b so S1 = S3*(c/b)².
factor ∆ACD → ∆BCF = a/b so S2 = S3*(a/b)².
S1 + S2 = S3*((c/b)² + (a/b)²) = S3*(a²+c²)/b²
Pythagoras in ∆ABC: a²+c²=b²
S1 + S2 = S3*b²/b² = S3.
QED.

(Note: there is a little error in the text on the side of the picture. F is the incenter of ∆BHC.)

3. Thanks Henkie for the note.

4. All 3 coloured triangles are similar having angles A/2, 45-A/2 and 135

Hence S1/c^2 = S2/a^2 = S3/b^2 = (S1+S2)/(c^2+a^2) = (S1+S2)/b^2 using Pythagoras

So S1 + S2 = S3

Here I've used the well known algebraic tool that if x/p = y/q then each ratio = (x+y)/(p+q)

Sumith Peiris
Moratuwa
Sri Lanka

5. Leave it to fellow Gogeometers to prove that AEFC is con cyclic

6. Also that D is the orthocentre of Tr. BEF