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

Post your solutions or ideas in the comments.

## Saturday, February 21, 2009

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

Labels:
congruence,
excenter,
isosceles,
right triangle

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Let B:(0,0), A:(0,a) and C:(a,0)

ReplyDeleteArea 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

which gives AD ~ 1.8477590650*a

while BE=a*cos(ABE)=acos(22.5)~ 0.923879532*a

2BE = 2*0.923879532*a = 1.847759065*a

Hence AD = 2BE

Ajit: ajitathle@gmail.com

from triangle BEA BE=AE*tan67.5, from triangle BED BE=(AE+AD)*tan22.5,then

ReplyDeleteAE=BEcot67.5 and AE=BE*cot22.5-AD. By using last two relations we get BE*cot67.5=BE*cot22.5-AD, then BE(tan67.5-tan22.5)=AD, BE(sin(67.5-22.5)\(cos67.5*cos22.5))=AD , BE(sin45\(sin22.5*cos22.5))=AD then AD=2BE

Shahlar Maharramov, Yasar University ,Izmir,Turkey and Cybernetics Institute Azerbaijan

http://img534.imageshack.us/img534/2876/problem19.png

ReplyDeleteBE cut DF at G.

Note that BFDH is a square >> BF=DF ( see picture)

Triangle BFG congruence to Tri. DFA (Case SAS)

So AD=BG

In Triangle GDB , ED is an angle bisector and ED is an altitude so BGD is isosceles

So BG=2BE=AD

Peter Tran

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

ReplyDeleteComplete 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

ReplyDeleteSo 2BE = AD

Sumith Peiris

Moratuwa

Sri Lanka

my solution here:

ReplyDeletehttps://www.facebook.com/photo.php?fbid=10206216363756695&set=a.10205987640598759.1073741831.1492805539&type=3&theater

Let F be the tangent point with AB and G the tangent point with AC and s be the length of AB.

ReplyDelete1. AG is sqrt(2)/2 * S.

2. tr ADF is congruent to ADH (2 sides and they are both right triangles) so AF = AG = sqrt(2)/2 * S.

3. DF and BF are both the radius length of the circle and = (1 + sqrt(2)/2) S

4. Using the Pythagorean theorem AD is therefore sqrt(2 + sqrt(2))

5. Since ABE is similar to ADF we can just use ratios to find BE = (1 / sqrt(2 + sqrt(2)) * ((2 + sqrt(2))/2) = sqrt(2 + sqrt(2))/2 = AD /2