tag:blogger.com,1999:blog-6933544261975483399.post4886479270512318541..comments2023-03-25T01:08:45.796-07:00Comments on Go Geometry (Problem Solutions): Problem 25: Right triangle, Altitude, Incircles, Incenter, Angle BisectorAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-72468501761905555652018-01-17T09:45:41.763-08:002018-01-17T09:45:41.763-08:00The first part:
From the point E draw a line perp...The first part:<br /><br />From the point E draw a line perpendicular onto the line BD. The line meets at the point Q. In problem 23 we have BD= r+r1+r2, therefore :<br /><br />BQ=BD-r=r1+r2 <br /><br />and from problem 28 we have <br /><br />EQ= r2-r1<br /><br />Draw a line from the point G perpendicular onto AC. The line meets at the point N.<br />Then draw a line from the point F perpendicular onto GN. The line meets at the point M. <br /><br />Then we have :<br /><br />FM=r1+r2<br /><br />and <br /><br />GM = r2-r1<br /><br />The triangles FMG and BQE are congruent from SAS and FG=BE<br /><br />Second part :<br /><br />We have : <br /><br />∠GFM =∠EBQ=u<br /><br />Let the line FG meet the line BD at the point P, then :<br /><br />∠FPD=∠BPG=90-u<br /><br />If we extend the line BE to meet FG at a point K and let : <br /><br />∠BKP=x <br /><br />than the angles of the triangle of BKP : <br /><br />180=∠BKP+∠EBQ+∠BPK<br />180=x+u+90-u<br />90=x<br />Arifhttps://www.blogger.com/profile/16774426807135540272noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23954873513597209182016-04-25T10:26:35.677-07:002016-04-25T10:26:35.677-07:00FG//EL and FG=EL
See my solution at https://www.fa...FG//EL and FG=EL<br />See my solution at https://www.facebook.com/photo.php?fbid=10206229971456879&set=a.10205987640598759.1073741831.1492805539&type=3&theaterLoukas Sidiropouloshttps://www.blogger.com/profile/05815628945558031099noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48427331879554367592015-10-22T22:14:07.127-07:002015-10-22T22:14:07.127-07:00FG^2 = (r1-r2)^2 + (r1+r2)^2 = 2(r1^2+r2^2) = 2r^2...FG^2 = (r1-r2)^2 + (r1+r2)^2 = 2(r1^2+r2^2) = 2r^2 by referring to my proof for Problem 24<br /><br />But BE^2 = 2r^2 by considering the square whose diagonal is BE<br /><br />Hence BE = FG <br /><br />For the 2nd part of this problem I salute Nilton Lapa's excellent solution above. <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-40956759493383849692012-03-06T15:23:17.317-08:002012-03-06T15:23:17.317-08:00I had the idea below, I’d like to know your opinio...I had the idea below, I’d like to know your opinion.<br />Let L on AB, and M on BC, be the tangent points of the incircle of ABC. Let P on AB be the tangent point of the incircle of ABD, and let Q on BC be the tangent point of the incircle of BDC. BLEM is a square with side r, which diagonals LM and BE are congruent and perpendicular. From problem 023, BD=r+r1+r2. As LP=BP-BL, BP=BD-r1, and BD=r+r1+r2 (this is the result of Problem 023), then LP=r2. Similarly, MQ=r1. So, LPF and GQM are congruent, with LF=MG. As LP and QG are paralel, and PF and MQ are paralel too, then LF and MG are paralel. This shows that LMGF is a paralelogram, with LM and FG congruent and paralel. But LM and BE are perpendicular, so BE and FG are congruent and perpendicular.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2829508099839113422012-03-04T22:33:49.907-08:002012-03-04T22:33:49.907-08:00http://img821.imageshack.us/img821/6514/problem25....http://img821.imageshack.us/img821/6514/problem25.png<br />Hope that my comment below will answer to questions of Peter and Nilton Lapa<br />Let r, r1, r2 are radii of incircles of triangles ABC, ADB and BDC ( see sketch)<br />1. Since these triangles are similar so r1= r. (AB/AC) and r2=r.(BC/AC)<br /> We have r1^2+r2^2= r^2.(AB^2+BC^2)/AC^2 <br />But AC^2=AB^2+BC^2 so r1^2+r2^2= r^2<br />BE^2= 2*r^2 and FG^2 =(r2-r1)^2+(r2+r1)^2 =2(r1^2 +r2^2)= 2.r^2<br />So BE=FG<br /><br />2. Let alpha= ∠ (ACB)= ∠ (ABD) and x= ∠ (KGH), y=∠ (KBE)<br />Since ADB similar to BDC so r1/r2=AB/BC= tan(alpha)<br />We have tan(x)=(r2-r1)/(r2+r1)= (1-tan(alpha))/(1+tan(alpha)) <br />Tan(y)= tan( m(ABE)-m(ABD))= tan(45-alpha)= (1-tan(alpha))/(1+tan(alpha)) <br />So x=y <br />Since ∠ (KHG) complement to x so ∠ (KHG) complement to y . <br />In triangle BHL, ∠ (KHG) complement to ∠ (KBE) so BE perpendicular to FGPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72438329603040134452012-03-04T14:00:24.540-08:002012-03-04T14:00:24.540-08:00To Bae Deok Rak
In (3), what are point X, Y? Witho...To Bae Deok Rak<br />In (3), what are point X, Y? Without defining their positions, it is impossible to understand the proof.<br />Please answer my question.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-24985085266898080772010-10-08T00:52:39.368-07:002010-10-08T00:52:39.368-07:00To Bae deo rak
It is not clear to me how do you g...To Bae deo rak<br /><br />It is not clear to me how do you get r^2 =r1^2+ r2^2 from r=(AB*BC)/(AB+BC+CA),r1=(AD*BD)/(AB+BD+DA) and r2=(CD*BD)/(BC+CD+DB)<br />Please explainPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45895976619345562612010-10-05T19:46:27.643-07:002010-10-05T19:46:27.643-07:00Let r, r1, and r2, are the inradii of the triangle...Let r, r1, and r2, are the inradii of the triangles ABC, ADB, and BDC respectively.<br />Then r=(AB*BC)/(AB+BC+CA),r1=(AD*BD)/(AB+BD+DA) and r2=(CD*BD)/(BC+CD+DB). Hence we get<br />(1) r^2 =r1^2+ r2^2.<br /><br />(2) Since BE=sqrt(2) *r and <br />FG^2 =(r2-r1)^2+(r2+r1)^2 =2(r1^2 +r2^2), <br />we get BE=FG.<br /><br />(3) Let X and Y be points on BA and BC, respectively.<br />Since two triangles ABD and DC are similar,<br />FD/XF=GD/YG, that is, XY // FG.<br />Now, we see that XY is perpendicular to BE and hence FG is perpendicular to BE.bae deok raknoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68646342581239863612009-04-08T23:41:00.000-07:002009-04-08T23:41:00.000-07:00Let incircles of Tr ABD & Tr.CBD meet AB &...Let incircles of Tr ABD & Tr.CBD meet AB & BC respectively in J & K. Now with the standard nomenclature we can show that: BD=ac/b,AD=c^2/b<BR/>Hence, r1=Inradius of Tr.ABD=(ac/b*c^2/b)/(ca/b+c^2/b+c) = ac^2/(b(a+b+c))<BR/>Likewise, r2=Inradius of Tr.CBD=(ac/b*c^2/b)/(ca/b+c^2/b+c) = ca^2/(b(a+b+c))<BR/>Now take BA & Bc as the x-axis and y-axis. It can be shown by elementary geometry that F is <BR/>[ac^3/(b(b-a)(a+b+c)), ac^2/b(a+b+c)] since BJ is r1*cot(B/2). Likewise, G is [ca^2/b(a+b+c),ca^3/(b(b-c)(a+b+c))]. By the distance formula which is nothing but using Pythagoras: FG^2 = (ac)^2*(a^+c^2-ab)^2/(b(a-b)(a+b+c))^2 +ac)^2*(a^+c^2-bc)^2/(b(b-c)(a+b+c))^2 <BR/>Now a^+c^2=b^2; thus, FG^2 = 2(ac/(a+b+c))^2<BR/>But r = Inradius of Tr. ABC = ac/(a+b+c)<BR/>Hence, FG^2 =2r^2 ---(1) while from Tr. BLE we've BE^2=BL^2+LE^2 where L is the point at which incircle of Tr. ABC touches AB. Also BL=LE=r. Thus, BE^2 =2r^2 ---(2)<BR/>By equations (1) & (2) we've, FG=BE=V2*r<BR/>Now slope FG = [ca^3/(b(b-c)(a+b+c)) - ca^2/b(a+b+c)]/[ca^2/b(a+b+c)-ac^3/(b(b-a)(a+b+c))]<BR/>= ac/(a+b+c)/ -[ac/(a+b+c)]= - 1 which means FG makes an angle of 135 deg with x-axis BA and since BE is the angle bisector of angle ABC (=90 deg) it makes an angle of 45 deg to the x-axis i.e. to BA which in turn means that BE is perpendicular to FG.<BR/>QED<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com