See complete Problem 66

Triangle, Excircle, Tangents, Geometric Mean. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 66

Labels:
excircle,
geometric mean,
tangent,
triangle

Subscribe to:
Post Comments (Atom)

idk how to prove this segment someone help me

ReplyDeletename P, OP meet AE

ReplyDeletePOE = A/2 (perpendicular sides)

POC = C/2 (POT = C, PCTO cyclic, T tg points to BC, OC

bisector of kite )

=>

COE = A/2 + C/2 (1)

OBD = (180 - B)/2 = 90 - B/2 = A/2 + C/2 (2)

(1) & (2) =>

COE = OBD (3)

E = D (4) (AED isoceles,AO bisector + altit)

from (3) & (4)

▲COE ~ ▲OBD

=>

(DE/2)/e = d/(DE/2)

DE = 2√d∙e

-------------------------------------------

can you descripe more where name P

DeleteP the point where circle meet AE, T the point where circle meet BC

ReplyDeleteO is the excentre and OA bisects < A so OD = OE

ReplyDeleteTr. s BOD and COE have as angles 90-A/2, 90-B/2 and 90-C/2 and are hence similar

So DO/d = e/OE and since DO = OE = DE/2, the result follows

Sumith Peiris

Moratuwa

Sri Lanka