tag:blogger.com,1999:blog-6933544261975483399.post269627335137437290..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 415: Right Triangle, Cevian, Angles, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-17859212389957950672023-12-09T23:16:36.868-08:002023-12-09T23:16:36.868-08:00[For easy typing, I use a instead of alpha]
<AB...[For easy typing, I use a instead of alpha]<br /><ABD=90-3a<br /><ADB=90+a<br />x=90-2a<br /><br />In triangle ABD<br />sin(90+a)/AB=sin2a/BD<br />AB/BD=sin(90+a)/sin2a--------(1)<br /><br />In triangle BDC<br />sinx/BD=sin3a/CD<br />CD/BD=sin3a/sinx---------(2)<br /><br />Since AB=CD, (1)=(2)<br />sin(90+a)/sin2a=sin3a/sinx<br />cosa/2sinacosa=sin3a/sin(90-2a)<br />cos2a=2sinasin3a<br />cos2a=cos2a-cos4a<br />cos4a=0<br />4a=90<br />2a=45<br />x=90-2a=45<br /><br />Marconoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63816304261993848432015-09-28T10:52:17.767-07:002015-09-28T10:52:17.767-07:00Let BH be an altitude and let BE be the angle bise...Let BH be an altitude and let BE be the angle bisector of < HBC so that BH and BE trisect < 3@. <br /><br />Hence < ABH = x and < ABE = x + @ = < AEB so AB = AE = DC which in turn leads us to the conclusion that AD = EC<br /><br />Hence AB = BC and x = 45.<br /><br />Sumith Peiris<br />Moratuwa <br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69585555002030047582014-02-22T17:01:54.039-08:002014-02-22T17:01:54.039-08:00E is on AC such that BD=BE. <ABD=90-3a so <B...E is on AC such that BD=BE. <ABD=90-3a so <BED=<BDE=90-a=<ABE making AB=AE and triangles ABE and CDB congruent leading to AE=CB. AB=AE=CB so x=45.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53224793113797095542010-01-08T02:55:42.148-08:002010-01-08T02:55:42.148-08:00by sine rule on BAD triangle and BDC triangleby sine rule on BAD triangle and BDC triangleUnknownhttps://www.blogger.com/profile/09666413807593087589noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47389716311899825932010-01-07T07:11:56.861-08:002010-01-07T07:11:56.861-08:00E on DC
Let
m(DBE)=2a
m(BDE)=m(BED)=90-a
m(ABE)=9...E on DC<br />Let <br />m(DBE)=2a<br />m(BDE)=m(BED)=90-a<br />m(ABE)=90-a<br />BD=BE<br />AB=AE<br />{AD=BA-DE<br />EC=DC-DE<br />BA=DC}<br />AD=EC<br />x=2a<br />4a=90<br />x=45Anonymousnoreply@blogger.com