tag:blogger.com,1999:blog-6933544261975483399.comments2020-06-01T19:22:33.495-07:00Go GeometryAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5970125tag:blogger.com,1999:blog-6933544261975483399.post-38499528050903393032020-06-01T08:32:23.545-07:002020-06-01T08:32:23.545-07:00ABE,BCF,ADF are congruent right triangles.
So <...ABE,BCF,ADF are congruent right triangles.<br /><br />So < BAE = < FBC = < FAD = p say. <br /><br />AE & BF are hence perpendicualar and AGFD is concyclic and hence < GAF = 90-2p & so < GFA = 2p = < ADG<br /><br />Therefore < GAD = < AGD = 90-p and so b = GD = AD = a<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13458588013357139412020-05-27T12:58:43.406-07:002020-05-27T12:58:43.406-07:00By drawing the common tangent at T we can easily d...By drawing the common tangent at T we can easily deduce that AC//BD<br /><br />C^2/d^2 = CT.CD/DT.CD = CT/DT = AT/BT (because AC//BD) = AT.AB/BT.AB = a^2/b^2 <br /><br />From which a/c = b/d<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka<br />Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47869688792908171142020-05-25T06:19:29.928-07:002020-05-25T06:19:29.928-07:00BG+GC+AH+HD=AE+EB+CF+FD
subtitute
BG+GC=EB+CF+2.PG...BG+GC+AH+HD=AE+EB+CF+FD<br />subtitute<br />BG+GC=EB+CF+2.PG-EP-PF<br />AH+HD=AE+FD+2.PH-EP-PFc.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7892623630583225712020-05-25T01:57:23.888-07:002020-05-25T01:57:23.888-07:00Let PE = e, PF =f, PG = g and PH = h
Now we can a...Let PE = e, PF =f, PG = g and PH = h<br /><br />Now we can apply the Pitot Theorem 5 times to the 5 inscribed quadrilaterals<br /><br />AE + h = AH + e …..(1)<br />EB + g = BG + e …..(2)<br />CF + g = CG + f …..(3)<br />FD + h = DH + f …..(4)<br />AB + CD = BC + AD ….(5)<br /><br />(1) + (2) + (3) + (4) - (5) gives us the desired result<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70165853161242708732020-05-24T13:32:14.748-07:002020-05-24T13:32:14.748-07:00PA/AD=PBA/ABD=PCA/ACD=(PBA+PCA)/(ABD+ACD)=(PBA+PCA...PA/AD=PBA/ABD=PCA/ACD=(PBA+PCA)/(ABD+ACD)=(PBA+PCA)/ABC<br />PB/BE=PBC/BEC=PBA/BEA=(PBC+PBA)/(BEC+BEA)=(PBC+PBA)/ABC<br />PC/CF=PBC/CFB=PAC/CAF=(PBC+PAC)/(CFB+CAF)=(PBC+PAC)/ABC<br />Therefore<br />PA/AD+PB/BE+PC/CF=(PBA+PCA)/ABC+(PBC+PBA)/ABC+(PBC+PAC)/ABC<br />=(2PBA+2PCA+2PBC)/ABC=2(PBA+PCA+PBC)/ABC=2ABC/ABC=2rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57425405088447386962020-05-24T08:57:57.381-07:002020-05-24T08:57:57.381-07:00the similarity in triBGF AND triBIC u applied but ...the similarity in triBGF AND triBIC u applied but how can we say GF is parallel to ICAnonymoushttps://www.blogger.com/profile/01883009105159371974noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62745152119623619052020-05-24T08:55:14.539-07:002020-05-24T08:55:14.539-07:00I am not getting thought proces similarityI am not getting thought proces similarityAnonymoushttps://www.blogger.com/profile/01883009105159371974noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8614941669213972042020-05-24T08:52:03.593-07:002020-05-24T08:52:03.593-07:00Sir may I know the book you refer for geometry the...Sir may I know the book you refer for geometry the way solved the question is outstandingAnonymoushttps://www.blogger.com/profile/01883009105159371974noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42792086050668504012020-05-18T05:10:04.727-07:002020-05-18T05:10:04.727-07:00Grab and pen and a piece of paper, and illustrate ...Grab and pen and a piece of paper, and illustrate accordingly.<br /><br />Add point E to the figure and connect it to line C. Do so in such a way that line CE is the angle bisector of angle BCD. Next, connect point D to point E. That is all for the illustrations. <br /><br />We can see that:<br /><br />> Triangles CDE and CBE are congruent because<br />~ They are share line CE <br />~ Line CD = line BC<br />~ Angle DCE = angle BCE<br /><br />> Triangles CDE and ADE are congruent because<br />~ Due to two equal opposite angles and equal adjacent lengths, quadrilateral ADCE is a concave kite.<br />~ They share side DE<br />~ This is a special case where SSA works, but only because they make up a kite. <br /><br />> Since triangles ADE, CDE and CBE are congruent, angle AED = angle CED = angle CEB.<br />> Angle AED = angle CED = angle CEB = 180 / 3 = 60<br /><br />Now, we solve the problem:<br /><br />Just focus on triangle CBE. The three angles in a triangle add up to 180.<br />x + a + 60 = 180<br />x + a = 120<br />x = 120 - aAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31908288441896287672020-05-16T22:30:28.584-07:002020-05-16T22:30:28.584-07:00the url you posted doesn't have any imagethe url you posted doesn't have any imageAnonymoushttps://www.blogger.com/profile/07338518257958446445noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78525900343684690482020-05-16T20:33:17.319-07:002020-05-16T20:33:17.319-07:00http://gogeometry.com/school-college/5/p1452-cycli...http://gogeometry.com/school-college/5/p1452-cyclic-quadrilateral-sangaku-japanese-theorem-geogebra-ipad.htm<br /><br />Denote [XYZ]= inradius of triangle XYZ<br />Per the result of problem 1452. For any cyclic quadrilateral ABCD we always have<br />[ABD]+[BDC]=[ADC]+[ACB]<br />For 2nd triangulation of problem 1480: <br />In quadrilateral BAFE, apply the result of problem 1452 we have<br />[AFB]+[BFE]= [AFE]+[ADE]+[ABD]-[BDE]…. (1)<br />Similarly, for quadrilateral BEDC we have<br />[BED]+[BDC]=[AED]+[ADC]+[ACB]- [AEB]….(2)<br />Add (1) to (2) and note that [ABD]+[AED]=[BDE]+[AEB]<br />We get [AFB]+[BFE]+[BED]+[BDC]=[AFE]+[AED]+[ADC]+[ACB]<br />So summation of all inradii of 2nd triangulation and 1st triangulation are equal= <br />With similar way we will have the summation of all inradii of 3rd triangulation and 1st triangulation are also equal<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28829182930841980982020-05-16T05:14:36.091-07:002020-05-16T05:14:36.091-07:00This solution is clearer when illustrated. Grab a ...This solution is clearer when illustrated. Grab a pen and a piece of paper.<br /><br />Available information:<br />> Angle ABD = 45<br />> Angle BDC = 135<br />> Angle BAC = x<br />> Angle ACB = 3x<br />> Angle ABC = 180 - 4x<br /><br />Firstly, create a duplicate of triangle ABC. Do so such that the new triangle partially overlaps, and shares the same base AC as triangle ABC. You will see that the new triangle has an unmarked point. Name it as point E. Name the intersection of line AE and line BC as point O.<br /><br />We can conclude that:<br />> Triangle AOC is isosceles.<br />> Triangles ABC and ACE have the same angles and lengths.<br />> Angle BAE = angle BCE = 3x - x = 2x<br /><br />Connect point D to point O, forming line DO. Connect point E to point D, forming line DE, if you have not done so. Connect line B to line E, forming line BE.<br /><br />We can see that:<br />> Angle AOC = 180 - 2x<br />> Angle AOB = angle COE = 180 - (180 - 2x) = 2x<br />> Triangle BAO is isosceles because angle BAO = angle AOB = 2x<br />> Triangle ECO is isosceles because angle ECO = angle COE = 2x<br />> Line AC is parallel to line BE.<br />> Triangles AOC and BOE are similar.<br />> Triangles AOC and BOE are isosceles.<br /><br />Now, create a duplicate of triangle BOE. Do it in such a way that the new triangle shares base BE with triangle BOE, but does not overlap it. There will be one unmarked point. Name that point as point F. <br />All illustrations are complete. Prepare for the conclusions afterwards, there are a lot of it. <br /><br />Now we solve the problem:<br /><br />> Quadrilateral BOEF is a rhombus as line BO = line EO = line BF = line EF.<br />> Line AB = line BF = line EF = line CE.<br /><br />(i) These show that the obtained figure is half of an equilateral (but not a regular) octagon.<br /><br />> Triangles BFE and BOE have the same angles and sides because they are identical.<br />> Angle AEF + angle BFE = 180 - 2x + 2x = 180 so lines BF and AE are parallel.<br />> Line AB = line EF<br />> Therefore, ABFE is an isosceles trapezium.<br />> This same test can also be used on quadrilateral BCEF. It is also a trapezium.<br />> Since two trapeziums can be formed on the same half, it has 3 or more angles on each half.<br /><br />(ii) This shows that the obtained figure is half of an octagon that has 6 or more equal angles. <br /><br />> The angle formed when two adjacent points are connected to the centre is always 45.<br /><br />(iii) The angle formed when two adjacent points are connected to the centre is always 45.<br /><br />Only one octagon fits all three conditions. <br />That is the regular octagon -- all sides are equal, and all interior angles are equal to 135.<br /><br />Therefore,<br />180 - 2x = 135<br />2x = 45<br />x = 22.5<br /><br />This may be long-winded and all, but it is the one of the only solutions that do not require one to draw conclusions without proof.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9116807527547117172020-05-14T08:49:19.616-07:002020-05-14T08:49:19.616-07:00< BLH = < BAH = < HEA = < BEL, so BE =...< BLH = < BAH = < HEA = < BEL, so BE = BL<br />Similarly BM = BF = BE = BL hence MEFL is concyclic with B as centre<br /><br />Therefore < MKH = < MLH = < MFE so EF // HK<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lankasumith peirisnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2913906790602971192020-05-10T12:30:21.494-07:002020-05-10T12:30:21.494-07:00https://photos.app.goo.gl/KQZZUh3UnhQKozyPA
Let u...https://photos.app.goo.gl/KQZZUh3UnhQKozyPA<br /><br />Let u= ∠ (HAB)= ∠ (HEA)<br />And v=∠ (KCB)= ∠ (CFK)<br />And w=∠ (HKM)= ∠ (HLM)<br />1. quad. AHBL is cyclic => ∠ (HAB)= ∠ (BLH)= u= ∠(BEL)<br /> EBL is isosceles and BE=BL<br />Similarly we also have BM=BF=BE<br />So M, E, F, G are concyclic with B as the center<br />2. In quad. MEFL we have ∠ (EFM)= ∠ (MLE)= w<br />In quad. HKLM we have ∠ (HKM)= ∠ (HLM)<br />So ∠ (EFM)= ∠ (HKM) => EF//HK<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8166868872512794442020-05-10T07:17:11.106-07:002020-05-10T07:17:11.106-07:00Solution sent by c.t.e.o :
1) Triangles EBM, BFL ...Solution sent by <a href="https://www.blogger.com/profile/16937400830387715195" rel="nofollow">c.t.e.o</a> :<br /><br />1) Triangles EBM, BFL isosceles => B is center of the circle passes through M, E, F, L<br />2) Ang EFM = Ang HKM => EF//HK<br />Sketch for P1479<br />https://photos.app.goo.gl/cDnnRqmh2su1pcW99Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74131965868777558862020-05-10T00:23:43.017-07:002020-05-10T00:23:43.017-07:00Grab a pen and a piece of paper, and illustrate ac...Grab a pen and a piece of paper, and illustrate accordingly. <br /><br />Firstly, let us calculate some angles:<br />Angle ABC = 180 - 20 - 20 - 10 - 30 = 100<br />Angle AOC = 180 - 20 10 = 150<br /><br />Extend line AO until it touches line BC. We will name the place they meet as point D.<br />So far, we can conclude:<br />> Angle ADB = 180 - 20 - 100 = 60. <br /><br />Add a point E on AC and connect the point to D, forming line DE, such that a duplicate of triangle ABD is formed. It should have intersected with CO. Now, name the intersection point as P. Connect point O and point E to form line OE. There's a lotta stuff you should know after this point, so brace yourself.<br />At this point we can conclude:<br />> Angle AED = angle ABC, angle BDA = angle EDA.<br />> Triangle COD is an isosceles triangle because angle COD = angle DCO. Line OD = line OC.<br />> Angle EDC = 180 - angle BDA - angle EDA = 180 - 60 - 60 = 60<br />> Angles ODE and CDE are equal.<br />> Triangles OED and DEC are congruent by SAS.<br />> Angle DEC = 180 - angle AED = 180 - angle ABD = 180 - 100 = 80<br />> Angle DEO = angle DEC = 80 by congruence.<br />> Angle OBD = angle DEO = 80 by congruence.<br /><br />x = angle OBD = 80.Anonymousnoreply@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-54924590236505453572020-05-08T14:36:17.745-07:002020-05-08T14:36:17.745-07:00Using results of Problem 1253, circumcircles of Tr...Using results of Problem 1253, circumcircles of Triangle HA1A2, HB1B2,HC1C2 and ABC, pass through point P. Since HP is common chord of 3 circles, its centers will lie on perpendicular bisectors of HP. Since Ma,Mb and Mc are also center of these triangles. <br />Ma,Mb and Mc will be collinear. Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7114062139003941882020-05-08T14:29:07.557-07:002020-05-08T14:29:07.557-07:00Here is another solution.
Let point Hb is reflect...Here is another solution. <br />Let point Hb is reflection of point H across AC. It is easy to see that point Hb lies on circumcircle of triangle EHJ. <br />Similarly point Ha,Hc reflection of point H across BC and AB will lie on circumcircle of Triangle DHK and Triangle FGH respectively. . <br />Let these circumcircles of Triangle EHJ and DHK instersect at point P. <br />Angle HPHb= 2xAngle HEJ angle HPHa= 180 -2xAngle HDK <br />We get Angle HbPHa = Angle HPHb - Angle HPHa = 2x(Angle HEJ + HDK) -180 = 2x(180-C)- 180 = 180 - 2C <br />Since Angle HbBHa = 180 - 2C, hence Ha,Hb,B and P are concyclic. Ha,Hb and B lie on circumcircle of Triangle ABC hence P also lies on same circle. <br /><br />Since Hc also lies on circum circle of Triangle ABC, We get Angle HcPHb = 2xAngle A<br />Angle HPHb=2x Angle HEJ, we get Angle HcPH = 2xA - 2xAngle HEJ = 2xAngle GFH<br />Since Angle HFHc is also 2xAngle GFH, points H,Hc,F and P are concyclic. <br />Hence Circumcircle of Triangle FGH also passes through P. <br />Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77555430735328189582020-05-08T03:46:16.156-07:002020-05-08T03:46:16.156-07:00From Ceva's theorem ( the power of P1 ) we get...From Ceva's theorem ( the power of P1 ) we get(AB1/CB1).(CA1/BA1).(BC1/AC1) = 1<br />Substitute BA1/BC1 = BC2/BA2 and so on ... => (AB2/CB2).(CA2/BA2).(BC2/AC2) = 1<br />c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2200168093090924502020-05-07T13:25:26.293-07:002020-05-07T13:25:26.293-07:00Let n=BA1 .BA2 = BC1.BC2
and m=AB1.AB2=AC1.AC2 ; ...Let n=BA1 .BA2 = BC1.BC2 <br />and m=AB1.AB2=AC1.AC2 ; p=CA1.CA2=CB1.CB2<br />1. Apply Ceva’s theorem for triangle ABC and point of concurrent P1<br />(A1B/A1C) * (B1C/B1A) *(C1A/C1B)=1<br /><br />2. Replace BA1=m/BA2 ; CA1=p/CA2 etc…. in above equation and simplify we have :<br />we get (A2B/A2C) (B2C/B2A)(C2A/C2B)= 1 .<br />Per Ceva’s theorem AA2,BB2, CC2 are concurrent at P2<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33535318652953374022020-05-07T12:18:48.665-07:002020-05-07T12:18:48.665-07:00ABCD is a rectangle not a square. So EBC and CFD a...ABCD is a rectangle not a square. So EBC and CFD are not necessarily congruentAnonymoushttps://www.blogger.com/profile/10965479773013510870noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44903347897940578832020-05-07T11:47:32.852-07:002020-05-07T11:47:32.852-07:00https://photos.app.goo.gl/fuXMGA8D5W6nwNzV8
Defin...https://photos.app.goo.gl/fuXMGA8D5W6nwNzV8<br /><br />Define points K, P, Q as midpoint of CF, DF, CD ( see sketch)<br />Note that T, K, P are collinear <br />Same as W,Q,P and N, Q, K<br />Consider triangle CDF and secant ABF, we have<br />EC/ED x BC/BF x AD/AF = 1…..(1)<br />But EC/ED= TK/TP ; BC/BF= WP/WQ ; AD/AF=NQ/NK<br />Replace it in ( 1) then we have TK/TP x WP/WQ x NQ/NK= 1<br />So N, W, T are colinear per Menelaus’s theorem <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15246779118192344352020-05-07T10:20:30.587-07:002020-05-07T10:20:30.587-07:00Triangles EBC and CFD are congruent . Draw S2 side...Triangles EBC and CFD are congruent . Draw S2 sidewards S1<br />See sketch for explanation<br />https://photos.app.goo.gl/JBqEyJmzAVfuJ3Yb6c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35838519932539342602020-05-05T11:05:47.596-07:002020-05-05T11:05:47.596-07:00Thanks Peter for your kind words.
Also thanks for ...Thanks Peter for your kind words.<br />Also thanks for sharing nice diagram, it provides more clarity. <br /><br />Regards<br />Pradyumna Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.com