Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Tuesday, July 30, 2019
Geometry Problem 1442 Triangle, Altitudes, Orthocenter, Perpendicular, Congruent Angles, Similarity, Measurement, iPad apps
Subscribe to:
Post Comments (Atom)
Since < BDE = A, < MBG = (90-A) - (90-C) = C-A = < MCB
ReplyDeleteIt follows that BM^2 = MG.MC = 10
So BM = sqrt10
Sumith Peiris
Moratuwa
Sri Lanka
Nice solution, Sumith.
ReplyDeleteThank u Ajit
Delete