tag:blogger.com,1999:blog-6933544261975483399.post82192987318318305..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 989: Arbelos, Semicircles, Diameter, Perpendicular, 90 Degree, Common External Tangent, Rectangle, Midpoint of ArcAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-80739353098213123842021-01-16T04:06:07.511-08:002021-01-16T04:06:07.511-08:00https://photos.app.goo.gl/Ps4iad2ZZNznvwP17https://photos.app.goo.gl/Ps4iad2ZZNznvwP17c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87095015335767452812020-02-24T09:14:44.968-08:002020-02-24T09:14:44.968-08:00Let GH, BD cut at X. Let O be the centre of AC and...Let GH, BD cut at X. Let O be the centre of AC and let OD cut GH at Y.<br />Let AB = 2a, BC = 2b and let GH = 2t,<br /><br />Now XG = XB = XH = t. Also GF^2 = (a-b)^2 + (a+b)^2 = 4ab = 4t^2 = BD^2 hence BX = FX = DX = GX = t. It follows that BFDG is a rectangle<br /><br />If < DAO = p, < DOB = p = < BXH<br />Therefore OYXB is concyclic and so OY is perpendicular to EG<br />It follows that EY = YF and hence ED = DF which is the result we need<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21824515324532701522017-01-21T01:41:35.233-08:002017-01-21T01:41:35.233-08:00Draw a circle with diameter BD. It meets (AB) in X...Draw a circle with diameter BD. It meets (AB) in X and (BC) in Y.<br /><br /><BXD=<BYD=90 deg by Thales' theorem. <br /><BXA=<BYC=90 deg by Thales' theorem.<br />So X lies on AD and Y lies on BD. <br />Thales again: <XDY = <ADC = 90 deg.<br />Now note that <BXY = 90 deg - <BDX = <BAX. <br />So XY is tangent to the (AB) at X and similarly tangent to (BC) at Y. So X=G and Y=H.<br /><br />By similarity the tangent to the (AC) at D is parallel to GH. Let M be the midpoint of {AC), then MD is perpendicular to EF, hence bisects EF and bisects angle EMF, and D is midpoint of arc EF.<br />FvLnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19653798855532559172014-11-27T10:23:45.034-08:002014-11-27T10:23:45.034-08:00Because <GAC=<FHC, AGHC is cyclic. Let AG an...Because <GAC=<FHC, AGHC is cyclic. Let AG and HC meet at D'. <D'AC+<D'CA=90, so D' must be on big circle. Furthermore, because D'G*D'A=D'H*D'C from cyclic property of AGHC, D' must also be on radical axis of 2 small circles, which is their common tangent. Therefore, D' coincides with D. It is easy to see that GDHB has 3 right angles from Thales theorem in each circle and must therefore be rectangle.<br /><GAC=<FHC=arcDF/2+arcFC/2=arcDE+arcFC/2. <br />This means that arcDF=arcDE so D is midpoint of arcEF.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-61509374656131118112014-03-08T19:27:20.206-08:002014-03-08T19:27:20.206-08:00Thank you for your comment on my solution.
PeterThank you for your comment on my solution.<br />PeterPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33991874028134257822014-03-06T14:40:28.390-08:002014-03-06T14:40:28.390-08:00I have a comment on the second part.
DO || HN, and...I have a comment on the second part.<br />DO || HN, and they are perpendicular to EF.<br />So DO bisects both chord EF and arc EF.Unknownhttps://www.blogger.com/profile/09741825454377078442noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58087390385401889132014-03-05T10:05:34.886-08:002014-03-05T10:05:34.886-08:00http://s25.postimg.org/vsrf9vrnj/Prob_989.png
Let...http://s25.postimg.org/vsrf9vrnj/Prob_989.png<br /><br />Let M, L, N, O are points shown on the sketch<br />1. We have MG⊥GH and NH⊥GH<br />Since LG=LB=LH=> L is the midpoint of GH<br />∠(GBH)= ½(∠GMB+∠BNH)= 90<br />And ∠ (MLN)=1/2(∠GLB)+ ∠BLH)=90<br />In right triangle MLN , BL^2=BM.BN<br />In right triangle ADC , DB^2=BA.BC= 4. BM.BN=> L is the midpoint of BD<br />And DGBH is a parallelogram with right angle ∠GBH => DGBH is a rectangle<br />2. Connect AG and CH<br />We have ∠ (AGB=∠ (BGD)= 90 => A,G,D are collinear<br />Triangles AMG and AOD are isosceles with common angle ∠GAM<br />So ∠ (AGM)= ∠ (ADO) => OD//MG => OD ⊥ EF<br />So D is the midpoint of arc EF<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com