Let L, M, N , P are points as shown on the sketch Observe that AL= NC= half perimeter of triangle ABC- BC P is the midpoint of arc AC => M is the midpoint of AC and LN In trapezoid ILND , MO is the mid-base => O is the midpoint of ID Triangle IOP similar to triangle IDE ..( case AA) Since O is the midpoint of ID so DE= 2 x OP= 2.R
Lets assume touch point of incircle and excircle are F and G, and Midpoint of AC is M. It is easy to see that FM=GM=(a-c)/2 Also IF,OM and DG are parallel to each other ( all are perpendicular to AC), hence O is midpoint of ID. B, I and E are collinear, join BE and let it intersects circumcircle at point H, and IH=HE, thus H is midpoint of IE. Consider triangle IED, O is midpoint of ID and H is midpoint of IE, Hence DE=2*OH, since OH=R, DE=2R.
Join BIE. Let it cut circle(O)at M. Angle ECI is a right angle. M is the midpoint of arc AMC. So OM bisects AC at right angles. We are done if we can show O is the midpoint of ID. Let X, Y, Z be the projections of I. O. D on AC respectively. It is easy to see that XY = b/2 - (s -a) = (c -a)/2 = YZ. IX, OY, DZ being //, OYM is a midline // to DE in Triangle IDE. Hence DE = 2 OM = 2R. N Vijaya Prasad Rajahmundry - INDIA.
IE meets arc AC at P. By incenter-excenter lemma, IE=2*IP. OP and DE both perpendicular to AC, so triangle OIP similar to DIE with ratio 1:2, so DE=2*OP=2R
http://s22.postimg.org/nia3yt2fl/pro_1207.png
ReplyDeleteLet L, M, N , P are points as shown on the sketch
Observe that AL= NC= half perimeter of triangle ABC- BC
P is the midpoint of arc AC => M is the midpoint of AC and LN
In trapezoid ILND , MO is the mid-base => O is the midpoint of ID
Triangle IOP similar to triangle IDE ..( case AA)
Since O is the midpoint of ID so DE= 2 x OP= 2.R
Lets assume touch point of incircle and excircle are F and G, and Midpoint of AC is M. It is easy to see that FM=GM=(a-c)/2 Also IF,OM and DG are parallel to each other ( all are perpendicular to AC), hence O is midpoint of ID.
ReplyDeleteB, I and E are collinear, join BE and let it intersects circumcircle at point H, and IH=HE, thus H is midpoint of IE.
Consider triangle IED, O is midpoint of ID and H is midpoint of IE,
Hence DE=2*OH, since OH=R, DE=2R.
Join BIE. Let it cut circle(O)at M. Angle ECI is a right angle.
ReplyDeleteM is the midpoint of arc AMC. So OM bisects AC at right angles.
We are done if we can show O is the midpoint of ID.
Let X, Y, Z be the projections of I. O. D on AC respectively.
It is easy to see that XY = b/2 - (s -a) = (c -a)/2 = YZ.
IX, OY, DZ being //, OYM is a midline // to DE in Triangle IDE.
Hence DE = 2 OM = 2R.
N Vijaya Prasad
Rajahmundry - INDIA.
Bring IK⊥AC,MN⊥AC, EL⊥AC(M is midpoint arc AC).Is <MIC=<ABC/2+<ACB/2 , <MCI=<MCA+<ACI=<ABC/2+<ACB/2.Therefore <MIC=<MCI.So MI=MC.But IC⊥CE.
ReplyDeleteThen <MIC+<MEC=90 and <MCI+<MCE =90 therefore MI=MC=ME.So MO//ED
Is IO=OD. Therefore ED=//2OM=2R.
IE meets arc AC at P. By incenter-excenter lemma, IE=2*IP. OP and DE both perpendicular to AC, so triangle OIP similar to DIE with ratio 1:2, so DE=2*OP=2R
ReplyDelete