tag:blogger.com,1999:blog-6933544261975483399.post5157147524605281307..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1245: Cyclic or Inscribed Quadrilateral, Circle, Arc, Midpoint, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-32400824204365983402016-09-01T13:19:12.064-07:002016-09-01T13:19:12.064-07:00Notice that GE_|_FH, thus EF^2+GH^2=FG^2+EH^2, whe...Notice that GE_|_FH, thus EF^2+GH^2=FG^2+EH^2, wherefrom x^2=27.<br /><br />Best regardsStan Fulgerhttps://www.blogger.com/profile/07420210614111479737noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82161773566215549282016-08-15T18:36:28.262-07:002016-08-15T18:36:28.262-07:00https://goo.gl/photos/Raksk92QKNNeK9wC6
Observe t...https://goo.gl/photos/Raksk92QKNNeK9wC6<br /><br />Observe that OE, OF , OG and OH perpendicular to AB, BC, CD and AD.<br />Draw diameter FF’<br />We have ∠ (EOF’)= ∠ (B) … ( both angles supplement to ∠ (EOF) )<br />∠ (HOG)= ∠ (B)…. ( both angles supplement to∠ (D)) <br />Triangles EOF’ congruent to HOG ( case SAS) => EF’= HG<br />In right triangle FEF’ we have EF^2+EF’^2= EF^2+HG^2= 36+16= 52= 4. Radius^2<br />Similarly FG^2+EH^2= 4. Radius^2= 25+x^2<br />So x^2= 52-25=27 and x=3.sqrt(3)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com