tag:blogger.com,1999:blog-6933544261975483399.post155595975375311306..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1356: Quadrilateral, Triangle, Angle, 30 Degrees, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger18125tag:blogger.com,1999:blog-6933544261975483399.post-27912343110959786122018-12-21T09:40:52.814-08:002018-12-21T09:40:52.814-08:00https://photos.app.goo.gl/qNBMmWbtA4FiGasX7
Stan ...https://photos.app.goo.gl/qNBMmWbtA4FiGasX7<br /><br />Stan has an excellent solution. See sketch in the above link for details<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-6627673912410569362018-12-21T06:31:39.773-08:002018-12-21T06:31:39.773-08:00Another Solution
Let BX be perpendicular to AC, X...Another Solution<br /><br />Let BX be perpendicular to AC, X on AC<br />Draw a perpendicular from A to CD to meet BX extended at Y and CD at Z.<br /><br />Since CXZY is concyclic, < AYB = 30 & so ABYD is concyclic <br /><br />< CAB = 60 & Tr. AXZ is equilateral & since AY = CY, Tr. ACY is also equilateral <br />Hence AZ = ZY & so AD = DY.<br /><br />Hence in concyclic ABYD, < DBY = alpha and so theta - alpha = 2.alpha which gives theta = 3.alpha<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-81058230462931887452018-12-21T00:14:25.882-08:002018-12-21T00:14:25.882-08:00Slight recasting of Stan’s solution with his kind ...Slight recasting of Stan’s solution with his kind permission <br /><br />Let O be the circumcenter of Tr. ACD. <br />Then Tr. OAD is equilateral<br />ADOB is a kite & < DBO = alpha<br />ABCO is also a kite and so theta - alpha = 2.alpha & the result followsSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30594909714027437602018-12-20T08:11:25.315-08:002018-12-20T08:11:25.315-08:00∠(IOF) is the central angle of ∠(IEF) since I, A, ...∠(IOF) is the central angle of ∠(IEF) since I, A, D,E,F is cyclic and O is the center<br />but ∠(IEF)=30 so ∠(IOF)=60<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18140993501067902992018-12-20T06:26:55.957-08:002018-12-20T06:26:55.957-08:00To Agashe,
I still do not understand how you conc...To Agashe,<br /><br />I still do not understand how you conclude that D is the ex-center of triangle ABE. Can you elaborate more, or post a diagram?Ercan Cemhttps://www.blogger.com/profile/17715448112482768479noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77296271495267408982018-12-19T09:54:41.610-08:002018-12-19T09:54:41.610-08:00Stan, can you elaborate on "O is reflection o...Stan, can you elaborate on "O is reflection of A over BD, hence BD is bisector of <ABO"Ercan Cemhttps://www.blogger.com/profile/17715448112482768479noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85584170810179773042018-12-19T09:23:30.069-08:002018-12-19T09:23:30.069-08:00Peter: Can elaborate on this line: (FEI)=30 => ...Peter: Can elaborate on this line: (FEI)=30 => central angle ∠ (AOI)= ∠ (IOF)=60 (Yes, ∠(FEI), but I do not see how that implies ∠ (IOF)=60 (F is not on the circle.)Ercan Cemhttps://www.blogger.com/profile/17715448112482768479noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25420786950111564342018-07-07T20:36:49.051-07:002018-07-07T20:36:49.051-07:00See that O, circumcenter of triangle ADC is reflec...See that O, circumcenter of triangle ADC is reflection of A in BD, which solves the problem.Stan FULGERhttps://www.facebook.com/stan.fulgernoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7042561331376563282018-07-06T22:27:23.392-07:002018-07-06T22:27:23.392-07:00Take O the circumcenter of triangle ACD; ABCO is a...Take O the circumcenter of triangle ACD; ABCO is a kite, thus BO is angle bisector of <ABC. Triangle ADO is equilateral and O is reflection of A over BD, hence BD is bisector of <ABO, done.Stan FULGERhttps://www.facebook.com/stan.fulgernoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56436952197207549842018-07-04T06:14:24.688-07:002018-07-04T06:14:24.688-07:00Given ∠ACD = ∠ADB = 30°, and BA = BC.
Prove that...Given ∠ACD = ∠ADB = 30°, and BA = BC. <br />Prove that ∠DBC = 3 times ∠ABD. <br /><br />For E on CD, BE bisects ∠ABC. Connect AE. BE intersects AC at F. Then AE = CE. ∠BEC = ∠AEB = ∠AED = 60°. If ∠ABC = 4x, then ∠ABE =∠CBE = 2x. Bisect ∠BEA with EN, which meets BD at I. Connect AI; and because ∠IDA = 30° = 60° / 2 = ∠IEA, then IEDA is cyclic. Then ∠IED = 90° = ∠DAI.<br /><br />Produce AI to meet BE at P. Consider quadrilateral PEDA:<br />∠APE = 360° -120° - 90° - ∠EDA = 150° - ∠EDA <br />= 60° + ∠DBE.<br /><br />Consider Δ ABF: ∠BCF = ∠FAB <br />= 90° - 2x; ∠AID = 60°; ∠BIA = 120°; ∠IAB = 60° - ∠ABD.<br />∠FAI = 90° - ∠APF = 30° - ∠DBE. ∠EDB = 60° - ∠DBE.<br />∠PAB = 60° + ∠ABD - 2x = 60° - ∠DBE = ∠IAB = 60° - ∠ABD. <br />Thus ∠ABD = ∠DBE = x. And ∠DBC = 3x = ∠ABD * 3.Michael in NJhttps://www.blogger.com/profile/05442943078092136495noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-80706740094392003362018-06-25T19:54:32.644-07:002018-06-25T19:54:32.644-07:00https://photos.app.goo.gl/cX2DnsWrXZg6DC4q9
Draw ...https://photos.app.goo.gl/cX2DnsWrXZg6DC4q9<br /><br />Draw angle bisector BE of ∠ (ABC)<br />This bisector will perpendicular to AC and meet AC at E<br />So ∠ (CEB)=60=∠ (AEB)<br />Draw angle bisector EI of ∠ (BEA) => ∠ (BEI)= ∠ (AEI)= 30= ∠ (ADI)<br />Since ∠ (ADI)= ∠ (AEI)= 30 So ADEI is a cyclic quadri. <br />Since ∠ (ADI)= ∠ (FEI)=30 => central angle ∠ (AOI)= ∠ (IOF)=60<br />Triangle BAO congruent to BFO ( case SAS) => BD is the angle bisector of ∠ (ABE)<br /> and ∠ (CBD)= 3. ∠ (ABD)<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72656031945344446582018-06-16T01:52:48.679-07:002018-06-16T01:52:48.679-07:00Reflect A in BD to create the point X. Then angle ...Reflect A in BD to create the point X. Then angle ADX is 60 degrees, and triangle ADX is equilateral. Hence angle AXD = 60 degrees. <br />Since angle ACD= 30 degrees, it must be on the circumference of the circle centred at X with radius XA. (Angle on circumference is half that at centre)<br />Therefore triangles AXB and CXB are congruent (SSS) so angle ABX = angle CBX<br />Hence 2alpha = theta - alpha and the required result follows immediately.Bobhttps://www.blogger.com/profile/05404608656691293105noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91622845166226017382018-06-15T23:42:07.300-07:002018-06-15T23:42:07.300-07:00If O is the circumcenter of triangle ACD then tria...If O is the circumcenter of triangle ACD then triangle OAD is equilateral and O is the reflection of A in BD, thus <DBE=<ABD and BE is perpendicular bisector of AC, done.Stan FULGERnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69796659150114288622018-06-04T07:18:22.513-07:002018-06-04T07:18:22.513-07:00Bisector angle ABC, intersect with CD at E,
Refl...Bisector angle ABC, intersect with CD at E,<br />Reflect triangle ABD along line BE, get triangle BCF; angle CBF =alpha; angle BFC = 30 degree;<br />Because angle AEB, angle BEC are 60 degree, so angle AED is 60 degree, so angle CEF is 60 degree;<br />So A,E,F on the same line;<br />Make equilateral triangle BCG, angle BEC = angle BGC = 60 degree;<br />So BCEG on the same circle;<br />Because angle GEB = angle AEB=60 degree, <br />So G, A,E,F on the same line;<br />So angle EGC = angle EBC = half angle ABC;<br />Also BG= CG, angle BFC=30 degree = ½ angle BGC;<br />So B,C,F on the same semicircle;<br />Angle FGC= 2x angle FBC = 2 alpha;<br />Angle FGC =angle EGC =half angle ABC =2 alpha;<br />So, alpha + theta = 2 x ( 2 alpha);<br />So theta= 3 alpha;<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47314250230504704072018-05-25T12:34:26.590-07:002018-05-25T12:34:26.590-07:00According to previous coomments and the sketch of ...According to previous coomments and the sketch of Peter<br />name P( BE meet circle), H (AP meet BD), L ( BD meet circle)<br />=> ang ALD = ang AED = 60°, ang LAP = LEP = 30° => AP perpendicular to BD<br />from conguence of right triangles BAH and BPH => BH bisectorc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33586281432454338182018-05-22T00:38:16.310-07:002018-05-22T00:38:16.310-07:00If you draw a circle passing through AB such that ...If you draw a circle passing through AB such that the chord AB subtends angle AEB/2 at other points on circle. It will intersect CE at 2 ex-centers of Tr. ABE. According to diagram it must be ex-center opposite to B, hence D must be ex-center of Tr. ABE. Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89331132014230157612018-05-20T15:24:30.428-07:002018-05-20T15:24:30.428-07:00To Agashe
https://photos.app.goo.gl/RKe0iP6Y79jH0...To Agashe<br />https://photos.app.goo.gl/RKe0iP6Y79jH0Atf1<br />In my opinion, the statement of line 4 “Consider Triangle ABE, ED is external bisector Angle AEB and ADB = Angle AEB/2 = 30 Deg.” Is not enough to conclude that D is the ex-center of triangle ABE. Please provide more details . See above for the sketch .<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89872627454547977662018-05-11T03:22:29.248-07:002018-05-11T03:22:29.248-07:00Let Angle B=4x, draw angle bisector BE of Angle B ...Let Angle B=4x, draw angle bisector BE of Angle B such that E lies on CD. Since ABC is isosceles Triangle, BE is also perpendicular bisector of AC. We get Angle BEC= Angle AEB = Angle AED = 60 Deg. Also Angle ABE = Angle CBE = 2x. <br /><br />Consider Triangle ABE, ED is external bisector Angle AEB and ADB = Angle AEB/2 = 30 Deg. Hence D must be ex-center of Triangle ABE and BD must be bisector of Angle ABE <br />Hence Angle ABD = Angle EBD = x, We get Angle DBC = 3x = 3.Angle ABD .Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.com