tag:blogger.com,1999:blog-6933544261975483399.post9210939803368746033..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Triangle with Equilateral triangles, Rhombuses, ParallelogramsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-55058538181750933482010-02-24T12:44:44.785-08:002010-02-24T12:44:44.785-08:00a) A'BA"C has
A"C//BA', A"...a) A'BA"C has<br /> A"C//BA', A"B//CA' ( ang 120°+60°)<br /> BC perpendicular to A"A' (A"A' altitude for equilat tr)<br /><br /> => A'BA"C is rhombus<br /><br />b) tr ABC "goes" to B"CA', AB"C', A"CB' (see P240)<br /><br />A'BC'B" has<br />ang C'B"A' = 360°- (C+60+A)=300+B-(A+B+C)<br />ang C'B"A' = 120+B<br /><br />ang C'BA' = 120+B<br />=><br />C'B"A' = C'BA' (1)<br /><br />ang BC'B" = 60 - B<br />ang BA'B" = 60 - B<br />=> <br />BC'B" = BA'B" (2)<br /><br />(1) & (2) =><br /><br />BC'B"A' paralelogram<br /><br />c) A"C'B"C has<br /><br />ang B"C'A" = 60 - (60 - B) - (60 - A)<br />ang B"C'A" = A + B - 60<br /><br />ang A"CB" = C - (C - 60) - (C - 60) = 120 - C<br />= 120 - (180 - B - A)= A + B - 60<br /><br />B"C'A" = A"CB" (1)<br /><br />ang C'B"C = C + 60<br />ang C'A"C' = C + 60<br /><br />C'B"C = C'A"C' (2)<br /><br />(1) & (2) =><br /><br />C'A"CB" paralelogram<br />------------------------------------------c .t . e. onoreply@blogger.com