Question

Mand N are two points on the circumference of a circle

centre O.
The straight line AMB is the tangent to the circle at M.
Angle MON = y
Prove that angle BMN = 1/2 y

Answer 1

Step-by-step explanation:

We can make use of the symmetry of the problem and properties of similar triangles.

Here's a "paragraph proof."

Draw line OP perpendicular to chord MN, such that P lies on line MB. Let the point of intersection of OP with chord MN be designated point Q. Line OP bisects angle MON, so has measure y/2.

Triangles OPM and MPQ are similar (by AA similarity; both right triangles contain angle OPM), so angle QMP ≅ angle MOP. As we observed above, angle MOP has measure y/2, so angle QMP has measure y/2.

Angle BMN is another name for angle QMP, so it has measure y/2.

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Additional comments

By the HL congruence postulate, the perpendicular bisector of chord MN divides triangle MON into two congruent right triangles. (OM≅ON, MQ≅NQ). That means ∠MOQ≅∠NOQ, so OQ bisects angle MON, as we say above.

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Another way to come to the desired conclusion is to recognize that for any point X on long arc MN, angle MXN will have measure y/2. (inscribed angle theorem) This remains true in the limit as X approaches M. As X approaches M, secant XM approaches tangent line MB. So, in the limit, angle BMN is y/2.