tag:blogger.com,1999:blog-6933544261975483399.post3832198985788858957..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 796: Triangle, Altitude, Measurement of AnglesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-59459624881463185412021-04-13T00:36:41.382-07:002021-04-13T00:36:41.382-07:00If AD extended meets BC at G, AHGE is concyclic so...If AD extended meets BC at G, AHGE is concyclic so ADGC is also concyclic<br /><br />< AHE = 90 = (x+9) + (2x+6)<br /><br />Therefore x = 25<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-66881691296444147562013-10-10T06:55:35.754-07:002013-10-10T06:55:35.754-07:00http://www.mathematica.gr/forum/viewtopic.php?f=20...http://www.mathematica.gr/forum/viewtopic.php?f=20&t=32532&p=150623Μιχάλης Νάννοςhttps://www.blogger.com/profile/02379101429577964881noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50072070999828621662012-08-21T02:17:24.998-07:002012-08-21T02:17:24.998-07:00Extend AD to meet BC at E. Each of the angles HAE,...Extend AD to meet BC at E. Each of the angles HAE, HBE being x, the 4 points H, A, B, E are con-cyclic. It then follows that angle BDA = BHA = 90. So D is the ortho-center of triangle ABC and CD is perpendicular to AB. Hence the sum of the angles BAE and ABE is 90 degrees, which means x + 9 + x + x + 6 = 90, 3x = 75, x = 25 degrees.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51203511891209182522012-08-19T19:04:07.426-07:002012-08-19T19:04:07.426-07:00Angle HDC = 2x+9
By comparing triangle ADH and DH...Angle HDC = 2x+9<br /><br />By comparing triangle ADH and DHC, we have HC/AH = tan(2x+9)*tan(x)<br />By comparing traingel ABH and HBC, we have HC/AH = tan(x) / tan (x+6)<br /><br />Therefore, tan(2x+9)*tan(x) = tan(x) / tan (x+6) => tan(2x+9)*tan(x+6) = 1<br /><br />Using product to sum, of sine and consine, we have <br />cos((2x+9)-(x+6)) - cos((2x+9)+(x+6)) = cos((2x+9)-(x+6)) + cos((2x+9)+(x+6))<br />cos((2x+9)+(x+6))=0 <br />3x + 15 = 90<br />x = 25.<br /> W Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89761726382305305122012-08-19T13:39:51.072-07:002012-08-19T13:39:51.072-07:00http://img13.imageshack.us/img13/6435/problem796.p...http://img13.imageshack.us/img13/6435/problem796.png<br /><br />Draw lines per attached sketch<br />Quadrilateral AHEB is cyclic => AE ⊥ BC<br /> Quadrilateral ACEF is cyclic => CF ⊥ AB<br />In AHEB angle BAE= ½(180-4x-12)=84-2x …….(1)<br />In ACEF angle BAE=x+9…… (2)<br />From (1) and (2) we have 84-2x=x+9 => x=25<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68076452366140167092012-08-19T11:36:35.098-07:002012-08-19T11:36:35.098-07:00Al prolongar AD intersecta a BC en ángulo recto po...Al prolongar AD intersecta a BC en ángulo recto por ser el ángulo C=90-x. Así D es el ortocentro del triángulo y debe cumplirse que: x+6+x+x+9=90 y por tanto x=25°.<br /><br />Migue.Anonymousnoreply@blogger.com