Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 719 details.
Monday, January 23, 2012
Problem 719: Incenter, Intersecting Circles, Angle, Measurement
Subscribe to:
Post Comments (Atom)
Equal arcs AO’, BO’ of circle (O) subtend equal angles at the circumference.
ReplyDeleteSo ∠ACO’ = ∠BCO’ and so
CD bisects ∠ACB …… (i)
Next ∠DAB =(1/2)∠DO’B referred to circle(O’)
=(1/2)∠CO’B =(1/2)∠CAB referred to circle(O) and so
AD bisects ∠CAB …… (ii)
From (i) and (ii), it follows that
D is the incentre of ∆ABC.
AO' = BO', so in cyclic quadrilateral AO'BC, CO' bisects <ACB.
ReplyDeleteMoreover if <ABC = B, then < AO'C = B (in Circle O) and so <ADO' = 90 - B/2
Hence < CAD = (90-B/2) - C/2 = A/2
Therefore DA bisects <A in Tr. ABC
Hence D is the incentre of Tr. ABC
Sumith Peiris
Moratuwa
Sri Lanka
I'm sure you are long gone, but how do you know that AO' and BO' are equal arcs without assuming that ray CO' bisects <C?
ReplyDelete