tag:blogger.com,1999:blog-6933544261975483399.post9191401172909790663..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1000. Scalene Triangle, Isosceles, Angle, 120 Degree, Midpoint, Distance, Equilateral, Congruence, Perpendicular, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-37992398754137472112021-04-20T14:19:02.902-07:002021-04-20T14:19:02.902-07:00Tr. ABE and CBD are congruent (SAS) => AE=8
Sin...Tr. ABE and CBD are congruent (SAS) => AE=8<br />Since G and H are mid-points of AC and EC => GH//AE and GH=4 -----(1)<br />Similarly GF//CD and GF=4 -----------------(2)<br />As ABD is isosceles=> m(BDF)=30 and Tr.DBF is 30-60-90 and BF=DB/2. <br />Similarly BH=EB/2<br />Tr. BFH similar to ABE with each side of BFH half of corresponding sides of ABE<br />Hence HF=AE/2=4 ---------(3)<br />From (1) ,(2) & (3) GFH is an equilateral triangle with each side = 4 units<br />=> FH=2Sqrt(3)Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-932892363058646672019-02-19T21:48:57.063-08:002019-02-19T21:48:57.063-08:00Following on from my proof of Problem 998, FGH is ...Following on from my proof of Problem 998, FGH is equilateral too and is of side 4 so<br />x = sqrt(16 - 4) = 2sqrt3<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76266377816860699162014-04-04T20:28:50.944-07:002014-04-04T20:28:50.944-07:00http://s22.postimg.org/iff0mclox/problem_1000.png
...http://s22.postimg.org/iff0mclox/problem_1000.png<br /><br />Let P is the midpoint of DE<br />As the result of problem 998 we have<br />Triangle DBC congruent to ABE=> AF=DC=8<br />And triangle PFH and FHG are equilateral<br />Since G and H are midpoints of AC and EC => GH= 1/2AE=4<br />In equilateral triangle FGH altitude FJ= GH.sqrt(3)/2= 2.sqrt(3)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63516218456040890992014-04-04T18:34:03.673-07:002014-04-04T18:34:03.673-07:00CD=AE=2GH=8, GH=4, JH=2.
By problem 998, FH=4.
H...CD=AE=2GH=8, GH=4, JH=2. <br />By problem 998, FH=4. <br />Hence, x=√12=2√3. <br />Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com