tag:blogger.com,1999:blog-6933544261975483399.post1633857476096627931..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 986: Triangle, Median, Midpoint, Equal Angles, 45 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-75870098100061692882015-06-28T06:54:54.444-07:002015-06-28T06:54:54.444-07:00CH the altitude of Tr.BCD = 1/√2 ofCD. But BC=√2.C...CH the altitude of Tr.BCD = 1/√2 ofCD. But BC=√2.CD from similar Tr.s. So BCH is a right Tr. with CH=1/2 BC. Hence x=30Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39560623497973597752014-02-28T22:20:45.126-08:002014-02-28T22:20:45.126-08:00/_ABC = /_ABD + /_DBC = (45-x) + x = 45°. Draw CE .../_ABC = /_ABD + /_DBC = (45-x) + x = 45°. Draw CE perpendicular to AB. /_BCE = 45° and thus E lies on the perpendicular bisector of BC while /_CEB=2*/_CDB and thus E is the circumcentre of Tr. CDB which makes /_DEC = 2x and EC=ED. Now D is the circumcentre rt. Tr. AEC; hence ED=DC. In other words, Tr. EDC is equilateral or 2x=60° or x=30°Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41541334615772052232014-02-23T08:04:27.891-08:002014-02-23T08:04:27.891-08:00Circumcenter O of triangle DCB has <DOB=2(180-&...Circumcenter O of triangle DCB has <DOB=2(180-<ACB)=2(45+x)=90+2x, so <OBD=45-x=<ABD and O is on AB . x is always less than 45 so <ACB is obtuse so O is on the same side of the line DB as segment AB. <br />P is on AC such that OP is perpendicular to DC. Therefore, CO^2=CP*CA=DC/2*DC*2=DC^2 because <COB=2*45=90, making triangle COD equilateral, 2x=60, x=30.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57849071517381551092014-02-19T17:45:08.437-08:002014-02-19T17:45:08.437-08:00Non-elementary proof:
ΔABC~ΔBDC
If DC=1, AC=2, t...Non-elementary proof: <br /><br />ΔABC~ΔBDC<br />If DC=1, AC=2, then BC=√2. <br /><br />By sine law, <br />sin x / 1 = sin 45° / √2<br />sin x = 1/2<br />x = 30°Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12831263511325907162014-02-19T17:37:20.027-08:002014-02-19T17:37:20.027-08:00http://s25.postimg.org/7verkimrj/pro_986.png
We ha...http://s25.postimg.org/7verkimrj/pro_986.png<br />We have ∆ (BDC) similar to ∆ (ABC ) …. Case AA<br />So ∠(ABC)=45<br />Draw circumcircle of ADB, BC will tangent to this circle<br />∠ (AOB)=2∠ (DBC)=90<br />From D draw a line //OA . This line will bisect OB => ∠ (DOB)=60<br />So x=∠ (DAB)= ½ ∠ (DOB)=30<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com