## Saturday, February 21, 2009

### Problem 19: Isosceles Right Triangle, Excenter, Perpendicular, Congruence

Proposed Problem
Try to use elementary geometry (Euclid's Elements)

See complete Problem 19 at:
gogeometry.com/problem/problem019.htm

Level: High School, SAT Prep, College geometry

1. Let B:(0,0), A:(0,a) and C:(a,0)
Area ABC=a^2/2, AC=V2*a where V=square root
Hence exradius =(a^2/2)/(2a -v2a) = a/(2-v2)
Hence AD^2 = (a/(2-v2))^2 + (a/v2)^2
while BE=a*cos(ABE)=acos(22.5)~ 0.923879532*a
2BE = 2*0.923879532*a = 1.847759065*a
Ajit: ajitathle@gmail.com

2. from triangle BEA BE=AE*tan67.5, from triangle BED BE=(AE+AD)*tan22.5,then
Shahlar Maharramov, Yasar University ,Izmir,Turkey and Cybernetics Institute Azerbaijan

3. http://img534.imageshack.us/img534/2876/problem19.png
BE cut DF at G.
Note that BFDH is a square >> BF=DF ( see picture)
Triangle BFG congruence to Tri. DFA (Case SAS)
In Triangle GDB , ED is an angle bisector and ED is an altitude so BGD is isosceles
Peter Tran

4. http://henrik-geometry.webs.com/19.htm

5. Complete the square BFDG and let DF & BE meet at H. Then Tr.s HFB & AFD are congruent SAA. So the 2 hypotenuses are equal. BE = EH since DE is both the altitude and the angle bisector

Sumith Peiris
Moratuwa
Sri Lanka

6. my solution here: