Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Monday, July 16, 2018
Geometry Problem 1365: Circle, Tangent, Secant, Chord, Parallel, Midpoint, Sketch, iPad Apps
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< AMB = < BFC < ACB = < AOB since ABOC is concyclic
ReplyDeleteHence ABMO is concyclic and so < AMO = < ABO = 90
Therefore M is the midpoint of DE
Sumith Peiris
Moratuwa
Sri Lanka
https://photos.app.goo.gl/BGn1ZyJuoHQCZUsQ7
ReplyDeleteConnect OA, OB, OC
Since AB and AC tangent to circle O => ∠ (BOA)= ∠ (COA)
We have ∠ (BFC)= ½. ∠ (BOC)= ∠ (BOA)
∠ (BFC)= ∠ (BMA)= ∠ (BOA) => MOAB is cyclic quad.
So ∠ (OMA)= ∠ (OBA) = 90 degrees => M is the midpoint of ED
hi , can you explain how you decide that MOAB is cyclic quad.
ReplyDeletethanks
in quadrilateral MOAB , we have ∠ (BMA)= ∠ (BOA)
Deleteso MOAB is cyclic