tag:blogger.com,1999:blog-6933544261975483399.post785364522494742633..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1174: Triangle, Quadrilateral, Double, Triple, Angle, Congruence, Excenter, Angle BisectorAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-83013973631526792872018-09-25T06:22:18.633-07:002018-09-25T06:22:18.633-07:00Sorry for the typos
Corrected m(APC)=m(PAB)+m(PBA)...Sorry for the typos<br />Corrected m(APC)=m(PAB)+m(PBA)=180-4(α+θ) -----------(3) to<br />"m(APC)=m(PAD)+m(PDA)=180-4(α+θ)" -----------(3)<br /><br />Now consider the triangle BAX to "Now consider the triangle BAO" Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87982676157530883852018-09-24T13:50:34.109-07:002018-09-24T13:50:34.109-07:00Join AD and form the isosceles triangle ABD
E be f...Join AD and form the isosceles triangle ABD<br />E be foot of the perpendicular from B to AD<br />Say P be the point of intersection of BE and DC<br />Join AP and since ABD is isosceles m(CAP)=m(BAC)=θ and also APD is isosceles<br />=> C is the incenter of triangle ABP ----------(1)<br />A bit of angle chasing leads us to the below<br />m(BPC)=m(EPD)=2(α+θ) -----(2)<br />m(APC)=m(PAB)+m(PBA)=180-4(α+θ) -----------(3)<br /><br />From (1), we have (2)=(3)<br />=>(α+θ)=30 ----------(4)<br />Now consider the triangle BAX<br />we have X=180-(90+θ/2)-α/2<br />=> X=75 degreesSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78268948704430490422015-12-27T09:55:43.477-08:002015-12-27T09:55:43.477-08:00http://s17.postimg.org/59873mej3/pro_1174.png
Let...http://s17.postimg.org/59873mej3/pro_1174.png<br /><br />Let angle bisector of angle ABD meet CD at F and AD at E<br />Observe that due to symmetry we have<br />∠(BAF)= ∠ (BDF)=2. Theta<br />And tri. AFD is isosceles<br />AC and BC are angles bisectors of ∠ (BAC) and ∠ (CFA) <br />So ∠ (BFC)= ∠ (CFA)=2.alpha+2.theta => ∠ (FDA)=alpha+theta<br />In right tri. BED we have alpha+theta=30<br />We have∠ (HOA)= theta/2 and ∠ (ABO)=alpha/2<br />In right tri. HBO we have x+theta/2+alpha/2= 90 => x= 75<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72571460807890387782015-12-27T01:47:17.002-08:002015-12-27T01:47:17.002-08:00Let the bisector of < ABC meet AC at E, CD at F...Let the bisector of < ABC meet AC at E, CD at F and AD at G.<br />For ease of typing I use @ for alpha and $ for theta <br /><br />Easily < ABO = OBC = @/2 and < CBE = @ so that < EBD = 2@<br />Also since BG is the perpendicular bisector of AD, < BAE = EAF = BDE = EDF = $<br />So C is the incentre of Tr. ABF so < DFG = < AFG = < AFC = < CFE and each of these must therefore be essentially 60 degrees since the last 3 of these 4 angles add upto 180.<br />Hence from Tr. BDF, 2(&+$) = 60 and so <br />&+$ = 30.....(1)<br /><br />In Tr. ABC, OA is an external angle bisector at A so x+@/2 = 90 -$/2 from which<br />x + (@+$)/2 = 90 <br />Substituting @+& = 30 from (1) therefore<br />x+30/2 = 90 and hence x = 75<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com