Geometry Problem

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

Click the figure below to see the complete problem 976.

## Saturday, February 8, 2014

### Geometry Problem 976: Isosceles Right Triangle, 45 Degrees, Incenter, Angle Bisector, Hypotenuse

Labels:
45 degrees,
angle bisector,
hypotenuse,
incenter,
isosceles,
right triangle

Subscribe to:
Post Comments (Atom)

Let D be a point on BC such that AD bisects ∠A.

ReplyDeleteThen AID is straight line.

∠BID=∠BAI+∠ABI=67.5°

∠BDI=90°−∠BAD=67.5°

Thus, BI=BD.

Hence, by Problem 975,

AC=AB+BD=AB+BI.

http://imageshack.com/a/img716/8509/6hr3.png

ReplyDeleteOn BC locate E such that BI=BE

Triangle IBE is isosceles with ∠ (IED)=22.5 degrees= ∠ (ICD)

Triangles IDC and IDE are congruent…. ( Case SAS) => DC=DE

And CA= 2 x CD=CE => CA=CB+BE= AB+BI

The angle bisector BI divides isoceles right Tr. ABC. into 2 congruent isoceles right Tr. s.

ReplyDeleteSo if the in radius be r,

AB - r = AC / 2 and BI + r = AC/2. Adding these 2 equations the result follows

Sumith Peiris

Moratuwa

Sri Lanka

Problem 976

ReplyDeleteLet point D in the extension of AB to B such that AD=AC.Then triangle ADI and ACI are congruent (AD=AC,AI=AI,<DAI=<CAI=22.5 ).So <BDI=<AC[=22.5 and <ABI=<BDI+<BID or

45=22.5+<BID or <BID=22.5 therefore BI=BD and AC=AD=AB+BD=AC+BI.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

(More of an algebric solution)

ReplyDeleteLet AB=BC=1 => AC = √2 and BI extended meet AC at P

Since m(BPC) = 90, triangle BPC is isosceles right angled

=> BP = CP = 1/√2

Semi-perimeter of ABC , S=(2+√2)/2

S.r=.5*AB.BC => r=IP=1/(2+√2)

=> BI = BP-IP = 1/√2 - 1/(2+√2)

=> BI = 2/2+2√2

=> BI = 1/1+√2 = √2-1

=> BI = AC-AB

=> AC = AB+BI

2r=a+b-c

ReplyDeletesqrt(2)BI=AB + BC - Ac =2AB-sqrt(2)AB

so sqrt(2) (BI+AB)=2AB=sqrt(2)AC

so BI+AB=AC