tag:blogger.com,1999:blog-6933544261975483399.post153502999830665774..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 999. Right Triangle, Midpoint, Median, Double Angle, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-6933544261975483399.post-72501885917635313822015-11-12T09:07:54.325-08:002015-11-12T09:07:54.325-08:00Alternatively from (3) above
4p^2 - r^2 = q^2
L...Alternatively from (3) above <br /><br />4p^2 - r^2 = q^2<br /><br />LHS = BC^2 from right Tr. BCD<br /><br />So BC = q and < BAC = 45 and hence 3x = 45 and x = 15Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-61965949057086821692015-11-12T05:44:19.350-08:002015-11-12T05:44:19.350-08:00H is the altitude of Tr. ABE drawn from EH is the altitude of Tr. ABE drawn from ESumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79523836332306399422015-11-11T23:12:37.302-08:002015-11-11T23:12:37.302-08:00Easily Tr. ABE is isoceles. Let AE = AB = q, DE = ...Easily Tr. ABE is isoceles. Let AE = AB = q, DE = EC = BE = p and BD = r<br /><br />Now BE is a tangent to circle ADE hence<br />p^2 = rq ......(1)<br /><br />Further AD is a tangent to circle ACE hence<br />(q-r)^2 = 2p^2 ....(2)<br /><br />From (1) and (2) we have <br />q^2 = 4p^2 -r^2 ......(3)<br /><br />Now consider h the altitude of Tr. ABE<br />h^2 = p^2 -r^2/4 = 4q^2 from .....,(3)<br /><br />So h = q/2 hence 2x = 30 and so x = 15<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69283197405856879012014-06-24T04:26:55.060-07:002014-06-24T04:26:55.060-07:00sorry Mr peter ((Using elementary geometry we can ...sorry Mr peter ((Using elementary geometry we can show that HA > HC and HN> HM<br />HN/HC > HM/HA so ∠(HCB)> ∠(HAB)))???????????Anonymoushttps://www.blogger.com/profile/05117313108449857467noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57726525861332001532014-06-03T11:05:13.216-07:002014-06-03T11:05:13.216-07:00Draw circle diameter AC, center F. B will be on th...Draw circle diameter AC, center F. B will be on this circle<br />Let assume that AHC is not isosceles and AB >BC<br />Point H is on radius FB. M and N are the projection of H over AB and AC<br />Using elementary geometry we can show that HA > HC and HN> HM<br />HN/HC > HM/HA so ∠(HCB)> ∠(HAB)<br />So If AB> BC , it doesn’t exist point H on FB so that ∠(HCB)= ∠(HAB) ( both angles are positive)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67506614542837678392014-05-30T15:27:45.804-07:002014-05-30T15:27:45.804-07:00why "∠BAH = ∠BCH implies ∆AHC isosceles."...why "∠BAH = ∠BCH implies ∆AHC isosceles." Thanks.????????Anonymoushttps://www.blogger.com/profile/14208563219141512109noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77301549020871146082014-05-22T14:42:28.823-07:002014-05-22T14:42:28.823-07:00as you AHC question why is isosceleyas you AHC question why is isosceleyAnonymoushttps://www.blogger.com/profile/14208563219141512109noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87859955601538084942014-05-19T21:22:32.172-07:002014-05-19T21:22:32.172-07:00Bleaug
Excellent solution.
Bleaug<br /><br />Excellent solution.<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10407718737294437152014-04-19T20:02:43.765-07:002014-04-19T20:02:43.765-07:00An equally challenging version of the above exerci...An equally challenging version of the above exercise (with solution) you can find here http://mfcosmos.com/archives/15173MATHS AND PHYSICS WORLDhttp://mfcosmos.comnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36689807620983867502014-04-12T15:06:26.273-07:002014-04-12T15:06:26.273-07:00A solution is here: http://mfcosmos.com/archives/1...A solution is here: http://mfcosmos.com/archives/15066MATHS AND PHYSICS WORLDhttp://mfcosmos.comnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67152355306482748232014-04-07T17:21:44.761-07:002014-04-07T17:21:44.761-07:00Please elaborate on why "∠BAH = ∠BCH implies ...Please elaborate on why "∠BAH = ∠BCH implies ∆AHC isosceles." Thanks.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8376281916356119922014-04-06T06:27:54.157-07:002014-04-06T06:27:54.157-07:00See http://bleaug.free.fr/gogeometry/999.png
BC p...See http://bleaug.free.fr/gogeometry/999.png<br /><br />BC perpendicular bisector FEG meets AC in F midpoint of AC, and E midpoint of DC (Thales). By construction ∆BEC and ∆BFC are isosceles. Right angle B implies ∆DEB is isosceles and similar to ∆BAE with acute vertex angle = 2x. <br /><br />Let H be the intersection of CD and BF. Since ∆BFC is isosceles, so is ∆BHE. Triangles ∆BAE and ∆BHE are isosceles and share the same base, hence AH bisects angle ∠BAE.<br /><br />∠BAH = ∠BCH implies ∆AHC isosceles. Hence 3x=45° or x=15°.<br /><br />bleaugAnonymousnoreply@blogger.com