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

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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