tag:blogger.com,1999:blog-6933544261975483399.post567755963029353334..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger46125tag:blogger.com,1999:blog-6933544261975483399.post-82634959552107336922022-07-24T17:32:54.574-07:002022-07-24T17:32:54.574-07:00Problem 1 Solution:
1. triangles ACB and BCD are...Problem 1 Solution:<br /><br />1. triangles ACB and BCD are similar: x = x, and <C = <C AA~ Theorem.<br />2. let AD = DC = a, let BC = b: a/b = b/(2a) so b/a = 1/sqrt2<br />3. Since the two triangles are similar, and m<BDC = 135 (linear pair with <ADB = 45), then m<ABC = 135. <br />3. Extend AB. Drop an altitude from C to line AB. Call the intersection E. m<EBC = 45.<br />4. Since m<AEC = 45 (altitude) and m<EBC = 45, triangle ECB is right isosceles: BE = EC. and BC is sqrt 2 times BE or EC.<br />5. Since, step 2, BC is sqrt 2 times DC, DC = BE = EC. Equilateral. m<ECD = m<CDE = m<DEC = 60. <br />6. Angle addition, 60 - 45 = 15 = m<BCD.<br />7. Triangle angle sum : 180 = 135 + 15 + x... x = 30.<br /><br />Mark Vasiceknoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51330907906625318802021-12-07T05:31:59.388-08:002021-12-07T05:31:59.388-08:00See the drawing
Let b=AD=DC and d=BC
ΔACB and ΔB...See the <a href="http://sciences.heptic.fr/archives/548" rel="nofollow"><b>drawing</b></a><br /><br />Let b=AD=DC and d=BC <br />ΔACB and ΔBCD are similar since ∠BAC=∠DBC=x and ∠BCD=∠BCA<br />Then AC/BC=CB/CD=AB/BD => 2b/d=d/b => d=b√2<br />Define C’ such as DC’ ⊥ AC and DC’=b<br />∠BDC=∠ADC - π/4= π - π/4 = 3 π/4<br />∠BDC’=∠BDA + ∠ADC’ = π/4 + π/2 = 3 π/4<br />=> ∠BDC= ∠BDC’<br />=>Δ BDC is congruent to ΔBDC’<br />=>BC=BC’=d and ∠DBC’=∠DBC=x => ∠C’BC=2x<br />DC’ ⊥ DC and DC’=DC=b => CC’ = b√2=d<br />ΔBCC’ is equilateral<br />∠C’BC = 60°=2x <br /><b>Therefore x=30°</b><br /><br /><br />rv.littlemanhttps://www.blogger.com/profile/05572092955468280791noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-81467619677450373272021-03-10T07:10:36.712-08:002021-03-10T07:10:36.712-08:00You can construct a circle passing through ANY thr...You can construct a circle passing through ANY three points A, B, C. The trick is to choose its centre O wisely. But this can be done for any triangle: just erect a perpendicular bisector on each of its sides. All three perpendicular bisectors should coincide in one and only one point – the circumcentre of the triangle (i.e. the centre of the circle that passes through all three of its vertices A, B, C).Anonymoushttps://www.blogger.com/profile/07091267898351713083noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42204793008215395962021-03-10T06:57:03.549-08:002021-03-10T06:57:03.549-08:00In order to assume that x is NOT 30°, one has to k...In order to assume that x is NOT 30°, one has to know first that it MIGHT BE 30°. This is putting the cart before the horse :qAnonymoushttps://www.blogger.com/profile/07091267898351713083noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31148610968427541812021-03-10T04:02:13.234-08:002021-03-10T04:02:13.234-08:00Most likely because of the fact that the two arcs ...Most likely because of the fact that the two arcs BC and CF are equal and adjacent. If you connect BF with a chord, this chord will be divided into two equal parts by the radius OC, which in this case will be perpendicular to that chord BF. (It wouldn't be the case if the arcs weren't equal.)Anonymoushttps://www.blogger.com/profile/07091267898351713083noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8363156025327228782021-02-10T01:27:51.485-08:002021-02-10T01:27:51.485-08:00Solution 2
Drop a perpendicular CE from C to AB e...Solution 2<br /><br />Drop a perpendicular CE from C to AB extended<br /><br />Easily Tr.BCE is right isosceles and so CE = a/sqrt2 .....(1)<br /><br />Since CB is tangential to Tr. ABD at B, a = b/sqrt2 ....(2)<br /><br />From (1) and (2) CE = b/2 and hence ACE is a 30-60-90 Triangle<br /><br />Therefore x = 30<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26710872938021949502020-12-08T15:16:53.682-08:002020-12-08T15:16:53.682-08:00Tr. ABC Similar to BDC
=>BC/AC=DC/BC
=>BC^2=...Tr. ABC Similar to BDC<br />=>BC/AC=DC/BC<br />=>BC^2=CA.CD ------(1)<br />Also given m(DBC)=m(BAD) -----(2)<br />From (1)&(2), CB is tangent to circumcircle of ABD. Let O be the center of this circle.<br />let AD=DC=a=> BC=a.Sqrt(2)<br />From the same similarity<br />AB/BC=BD/DC=>AB=BD.Sqrt(2)--------(3)<br />Since m(BDA)=45=>m(BOA)=90=>AOB is 45-90-45 triangle=>AB=BO.Sqrt(2)----(4)<br />From(3)&(4) OB=BD=Radius of the circumcircle.<br />However OB=OD=> BOD is an equilateral triangle <br />m(OBD)=90-x=60=>x=30 or m(BOD)=2.m(BAD)=2x=60=>x=30 degreesSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76273383491418902072020-05-09T20:06:29.724-07:002020-05-09T20:06:29.724-07:00Take a pen and a pice of paper, and illustrate acc...Take a pen and a pice of paper, and illustrate accordingly.<br /><br />Label angle BCD as y. Now, x + y = 45. <br />Label angle ABD as z. Now, x + z = 135.<br /><br />Firstly, add a point E and connect it to point A with line AE, and to point B with line BE, such that a duplicate of triangle ABD is formed. Since triangles ABE and ABD are identical, we can conclude that AD = AE, BD = BE.<br /><br />Next, add a point F and connect it to point A with line FA, and to point B with line FB, such that a duplicate of triangle BCD is formed. Since triangles BCD and BCF are identical, we can conclude that CF = CD, BF = BD. <br /><br />Now, we connect point E to point F with line EF. This is where the magic happens!<br /><br />Angle EBF will be:<br />360 - x - y - x - y = 360 - 135 - 135 = 90<br /><br />BD = BE<br />BD = BF<br /><br />Therefore, triangle EBF is an isosceles right angled triangle. This means that angle BEF = angle BFE = 45. <br /><br />Angle BFE + angle BFC = angle BFE + angle BDC = 45 + 135 = 180. This shows that E, F and C are collinear. Angle AEC = angle AEB + angle BEC = angle ADB + angle BEF = 45 + 45 = 90. <br /><br />Line AE = line AD. Line AC = 2 of line AD. Angle AEC = 90. These are properties of a 30-60-90 triangle. <br />Therefore, 2x = 60, and x = 30. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23957097665849322922020-03-25T10:35:36.321-07:002020-03-25T10:35:36.321-07:00Do you mean circumscribed circle of ABC
or BCD?Do you mean circumscribed circle of ABC<br />or BCD?Anonymoushttps://www.blogger.com/profile/11290084516238860097noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8925391950970754912019-05-12T07:42:31.314-07:002019-05-12T07:42:31.314-07:00What is the justification for (3)? (CO is perpendi...What is the justification for (3)? (CO is perpendicular to BF, intersecting at point E.)Ercan Cemhttps://www.blogger.com/profile/17715448112482768479noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76521678819094689012019-04-22T06:14:04.296-07:002019-04-22T06:14:04.296-07:00on the fourth line of the solution it was possible...on the fourth line of the solution it was possible to recall the sine theorem)Anonymoushttps://www.blogger.com/profile/13952594108039277104noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-759874281001072982018-12-27T09:16:20.595-08:002018-12-27T09:16:20.595-08:00see video : Problem 1 Solution Asee video : <a title="Problem 1 Solution A" href="https://youtu.be/EtQkUjrNubQ" rel="nofollow">Problem 1 Solution A</a>rv.littlemanhttps://www.blogger.com/profile/05572092955468280791noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67769234044651062582017-10-18T19:37:54.850-07:002017-10-18T19:37:54.850-07:00 Let E, be a point on (AD) ̅ such that (BE) ̅⊥(BD)... Let E, be a point on (AD) ̅ such that (BE) ̅⊥(BD) ̅.<br />Let F, be a point on (AD) ̅ such that (BF) ̅⊥(AD) ̅.<br /> m∠EBD=m∠BFD=m∠BFE=90°<br /> ∆EBD,∆BFD and ∆BFE are isosceles right triangles.<br /> BF=EF=FD=a, FA=b<br /> AD=CD<br /> AD=DC=AF+FD=a+b, AE=AF-EF=b-a<br /> BD=BE=√2 a<br /> m∠BEF=m∠EBF=m∠FBD=45°<br /> m∠AEB=m∠BDC=135°<br /> ∆AEB~∆BDC<br /> (√2 a)/(a+b)=(b-a)/(√2 a), 2a^2=b^2-a^2,3a^2=b^2, then b=√3 a<br /> c^2=a^2+b^2=a^2+(√3 a)^2=4a^2, c=2a<br />∆ABF is a 30°-60°-90° triangle with m∠A=x=30°<br />Anonymoushttps://www.blogger.com/profile/13757592567870837910noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7551314162830694992016-10-15T08:52:58.762-07:002016-10-15T08:52:58.762-07:00https://www.youtube.com/watch?v=I1wHTSSUW-whttps://www.youtube.com/watch?v=I1wHTSSUW-wgeoclidhttps://www.blogger.com/profile/07989522895596673545noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13566263673506658032016-04-27T07:37:04.273-07:002016-04-27T07:37:04.273-07:00The circumscribed circl with triangle BCD intersec...The circumscribed circl with triangle BCD intersects the perpendicular bisector AC in E. Then AE=EC and <DAB=<DBC=<DEC=x. If AB intersects EC in K so <DAK=<DEK=x .Is BK perpendicular EC(<KBC=<KAC+<BCD=<DBC+<BCD=45=<KBE ),then BC=BE, EK=KC. But triagle AKC=triangleAKE,so AE=AC <BAC=x=<BAE and triangle AEC it is equilateral. Therefore 2x=60<br />Or x=30.<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74466039824271899562016-04-18T10:14:17.251-07:002016-04-18T10:14:17.251-07:00https://www.facebook.com/photo.php?fbid=1020618961...https://www.facebook.com/photo.php?fbid=10206189617728061&set=a.10205987640598759.1073741831.1492805539&type=3&theaterLoukas Sidiropouloshttps://www.blogger.com/profile/05815628945558031099noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60636173994501567302015-09-27T14:01:39.167-07:002015-09-27T14:01:39.167-07:00The symmetrical of AC about AB and BC respectively...The symmetrical of AC about AB and BC respectively intersect at E; since <BAC+<BCA=45, we infer <AEC=90 and B is the incenter of triangle AEC, hence <AEB=45=<ADB, consequently triangles AEB and ADB are congruent (a.s.a criterion), making AE=AD. From right-angled triangle AEC we get DE=AD, triangle ADE equilateral and <EAD=60, consequently <BAD=30Stan Fulgerhttps://www.facebook.com/stan.fulgernoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90048533005846950882015-08-14T05:08:10.191-07:002015-08-14T05:08:10.191-07:00Since Tr.s ABD & BDC are equal in area, BD/sqr...Since Tr.s ABD & BDC are equal in area, BD/sqrt2 X AD = DE X BC where DE is the altitude. But since BC is tangential to Tr. ABD, BC = sqrt 2 X AD from which we deduce that DE = 1/2 BD and so x = 30<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90086709639854330412015-04-30T14:25:56.551-07:002015-04-30T14:25:56.551-07:00How do you presume point B lies on Circle O?How do you presume point B lies on Circle O?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79280911731165100322015-01-03T15:26:17.202-08:002015-01-03T15:26:17.202-08:00Take P - symmetrical of C about AB; clearly tr. BC...Take P - symmetrical of C about AB; clearly tr. BCP is right-angled and isosceles, hence <BPC=45=<ADB and BDCP is cyclic, so PD_|_AC, i.e. PC=AP, but by symmetry AP=AC and ACP is an equilateral triangle, or <BAC=30 degs. Best regards, Stan FulgerAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65284190713730831542013-10-31T09:38:43.198-07:002013-10-31T09:38:43.198-07:00corect link http://ogeometrie-cip.blogspot.ro/2012...corect link http://ogeometrie-cip.blogspot.ro/2012/08/problema-1.htmlcip1703https://www.blogger.com/profile/10825540554270680083noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16077534962803600672013-06-19T11:12:31.088-07:002013-06-19T11:12:31.088-07:00This is a very good problem
here is the solution h...This is a very good problem<br />here is the solution hope you understand it<br />Stratergy:In this problem we need to show that Tri BAEis 30-60-90(BE IS THE ALTITDE)for this we will use the thereom if one side of an right angeled Tri is half the hypotenuse the opp Ang is =30<br />PROOF<br /> TriBDC and TriABC are similar<br />BC/AC=DC/BC=BD/AB-----1<br />considerDC/BC=BC/AC<br />let DC=a<br />then BC=sq<br />CONSTRUCT Tri BED(BE IS THE ALTITDE)<br />LET BD=y<br />Th EB=ED=y/sqrt2---2<br />Using 1 <br />BA/BD=BC/DC<br />sqrt2*y/y=AB/y<br />Th sqrt2*y=AB-----3<br />Using the Stratergy<br />if BE*2=AB<br />THE PROOF IS COMPLETE<br />USING 1,2,3TO FILL VALUES WE GET<br />Y/sqrt2*2=BE<br />sqrt2*sqrt2*x/sqrt2=BE(2=sqrt2*sqrt2)<br />simplifying we get<br />BE=sqrt2*x<br />AB=sqrt2*x<br />AB=BE<br />Using Stratergy<br />x=30<br />Hence proveD<br /><br />Anonymoushttps://www.blogger.com/profile/00908049546546731829noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37997415006931806082013-02-18T04:09:15.873-08:002013-02-18T04:09:15.873-08:00Let AD=DC=a
Since BDC and ABC are similar BC=V2a
n...Let AD=DC=a<br />Since BDC and ABC are similar BC=V2a<br />now arBAD = arBDC or 1/2. a. BD. sin 45 = 1/2. BD. V2 a. sin x<br />sin x = 1/2 since x is acute so x = 30 degrees<br />kalpithttps://www.blogger.com/profile/08431159710282150473noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30647870773267748462012-12-10T03:33:24.181-08:002012-12-10T03:33:24.181-08:00Let AD = DC = t.
As seen above, [BDC] and [ABC] ar...Let AD = DC = t.<br />As seen above, [BDC] and [ABC] are similar => BC = sqrt(2) t.<br /><br />Drop a perpendicular from C to AB produced cutting it at E and let BE = s.<br /><br />By considering angle(EBD) as an exterior angle to [ABD], we have angle(EBC) = 45 degrees.<br />EBC is then an 45–45–90 triangle with EC = s and BC = sqrt(2) s<br />Thus, s = t<br /><br />In [AEC],<br />EC = t, AC = 2t and angle(E) = 90 degrees => AEC is an 30–60–90 triangle.<br />Therefore, x = 30 degrees<br /><br />Mike from Canotta<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23407761589838834842012-09-08T08:08:27.541-07:002012-09-08T08:08:27.541-07:00angle BDC = 180 - angle BDA
...angle BDC = 180 - angle BDA <br /> = 180 - 45 = 135 degree<br />we also know that angle BDC = angle DAB + angle ABD<br /> = x + angle ABD<br />ie angle ABD = (135-x)degree<br />now AD/sin(angle ABD)= BD/sin x<br />or BD/AD = sin x/ sin(135-x)........... 1<br />for triangle BDC<br />DC/sin(angle DBC)=BD/sin (angle BCD)<br />BD/DC= sin (angle BCD)/sin(angle DBC) = sin(45-x)/sin x...............2<br />from 1 and 2 we get <br />BD/AD=BD/DC [as AD=DC]<br />sin x/ sin(135-x)=sin(45-x)/sin x<br />or (sin x)^2 = sin(135-x)sin(45-x)<br />or (sin x)^2 = 1/root2(cos x + sin x) . 1/root2 (cos x - sin x)<br />or (sin x)^2 = 1/2[ (cos x)^2- (sin x)^2]<br />or 2.(sin x)^2=[ (cos x)^2- (sin x)^2]<br />or 3.(sin x)^2=(cos x)^2<br />or (tan x)^2 = 1/3<br />or tan x =1/root3<br />or x=30 degree<br />daily life of a middle class indianhttps://www.blogger.com/profile/06196646295163511134noreply@blogger.com