See complete Problem 66
Triangle, Excircle, Tangents, Geometric Mean. Level: High School, SAT Prep, College geometry
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Monday, May 19, 2008
Elearn Geometry Problem 66
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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
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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