tag:blogger.com,1999:blog-6933544261975483399.post4496528834320354097..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1373: Isosceles Triangle, Exterior Cevian, Inradius, Exradius, Altitude to the BaseAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-2887452439293934482018-10-16T11:24:08.576-07:002018-10-16T11:24:08.576-07:00Let BA=BC=a, UB=a1 and BV=a2 and r1 touch AB at Z
...Let BA=BC=a, UB=a1 and BV=a2 and r1 touch AB at Z<br />then UB=BZ=a1 => AG=a-a1, ||ly BV=BY=a2 => CH=a-a2 -------(1)<br />But UV=GH (since DV=DH)<br />=> a1+a2=GA+AE+EC+CH<br />=> a1+a2=a-a1+b+b+a-a2<br />=> a1+a2 = a+b<br />=> GH = a+b--------(2)<br /><br />Now GE= GH-EH<br />=> GE=(a+b)-(b+a-a2)<br />=> GE=a2=BYSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4204290570625324242018-10-14T09:48:28.595-07:002018-10-14T09:48:28.595-07:00How do we know BY=GE?How do we know BY=GE?mehmetnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45364997113740256772018-10-03T21:43:20.695-07:002018-10-03T21:43:20.695-07:00Is the reverse true? I mean, if the height of the ...Is the reverse true? I mean, if the height of the triangle is the sum of radii of the circles; is it always an isosceles triangle? I couldn't prove this. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21312978199620440102018-08-14T05:51:16.536-07:002018-08-14T05:51:16.536-07:00Correction :
let m(ABP)=@ instead of m(BPA)=@ Correction :<br />let m(ABP)=@ instead of m(BPA)=@ Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91789757120042085092018-08-13T11:13:57.509-07:002018-08-13T11:13:57.509-07:00Denote center of small circle as P and large circl...Denote center of small circle as P and large circle as Q<br />Drop a perpendicular from P to BE,denote it as X and form a right-triangle BPX<br />From Q, drop perpendicular to BC,denote it as Y and form a right-trianlge BQY<br />We will prove that these two triangles are congruent and the result follows<br /><br />Let the foot of perpendicular from P to DA be G and similarly from Q to DC be H<br />Similarly the foot of Perpendicular from P to DB be U and Q to DB be V<br />Let AB=BC=a , AE=EC=b, GA=x, CH=y<br />We know BV=BY=GE=PX=x+b ---------------(1)<br /><br />let m(BPA)=@ and m(BAC)=$<br />=>m(XBP)=90-$+@ and m(BPX)=$-@ ------------(2)<br />Similarly m(QBY)=$-@ and m(BQY)=90-$+@-------------(3)<br /><br />From (1),(2) and (3), BPX and QBY are congruent and BX=QY=r2<br />Since XE=r1 <br />=>h=BX+XE=r2+r1 Q.E.DSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.com