tag:blogger.com,1999:blog-6933544261975483399.post1324713248306118249..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 634: Right Triangle, Altitude, Incenters, AreasAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-35765263226099681072016-04-15T23:04:47.124-07:002016-04-15T23:04:47.124-07:00Also that D is the orthocentre of Tr. BEFAlso that D is the orthocentre of Tr. BEFSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69542404408642749692016-04-14T00:41:37.152-07:002016-04-14T00:41:37.152-07:00Leave it to fellow Gogeometers to prove that AEFC ...Leave it to fellow Gogeometers to prove that AEFC is con cyclic Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5820765990389152482016-04-14T00:26:28.277-07:002016-04-14T00:26:28.277-07:00All 3 coloured triangles are similar having angles...All 3 coloured triangles are similar having angles A/2, 45-A/2 and 135<br /><br />Hence S1/c^2 = S2/a^2 = S3/b^2 = (S1+S2)/(c^2+a^2) = (S1+S2)/b^2 using Pythagoras <br /><br />So S1 + S2 = S3<br /><br />Here I've used the well known algebraic tool that if x/p = y/q then each ratio = (x+y)/(p+q)<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72010133929001719992011-07-21T06:50:02.713-07:002011-07-21T06:50:02.713-07:00Thanks Henkie for the note.Thanks Henkie for the note.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31255636665910066282011-07-21T01:46:39.791-07:002011-07-21T01:46:39.791-07:00Call α = angle A, AB = c, BC = a and AC = b.
Becau...Call α = angle A, AB = c, BC = a and AC = b.<br />Because AE, BF and CD are bisectors, it is easy to see that ∆AEB, ∆BFC and ∆ACD all have angles ½α, 45º-½α and 135º, so they are all similar triangles.<br />factor ∆ACD → ∆ABE = c/b so S1 = S3*(c/b)².<br />factor ∆ACD → ∆BCF = a/b so S2 = S3*(a/b)².<br />S1 + S2 = S3*((c/b)² + (a/b)²) = S3*(a²+c²)/b²<br />Pythagoras in ∆ABC: a²+c²=b²<br />S1 + S2 = S3*b²/b² = S3.<br />QED.<br /><br />(Note: there is a little error in the text on the side of the picture. F is the incenter of ∆BHC.)Henkiehttps://www.blogger.com/profile/04279523252566471532noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36471754045371834612011-07-15T22:58:49.278-07:002011-07-15T22:58:49.278-07:00Note that Triangles ABC, BHC and ABH are similar t...Note that Triangles ABC, BHC and ABH are similar triangles ( Case AA)<br />In each triangle , ratio of height from 90 degrees angle/ radius of incenter circle = h1/r1= h2/r2=h3/r3<br />So ratio S(AEB)/S(AHB)=S(BFC)/S(BHC)=S(ADC)/S(ABC)<br />But we have S(ABC)=S(ABH)+S(BHC)<br />So S(AEB)+S(BFC)=S(ABC)<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com