Lines ma and mb tangent circle k(o) at a and b. point c is symmetric to point o with respect to point b . prove: m∠amc=3m∠bmc.

See explanation

Step-by-step explanation:

If MA is tangent to the circle k(O), then radius OA is perpendicular to segment MA.

If MB is tangent to the circle k(O), then radius OB is perpendicular to segment MB.

Consider two right triangles MOA and MOB. In these triangles:

Thus, triangles MOA and MOB are congruent by HL theorem. So

∠AMO=∠BMO.

If point C is symmetric to point O with respect to point B, then OC⊥MB. Consider two right triangles MOB and MCB. In these triangles:

Thus, triangles MOB and MCB are congruent by HL theorem. So

∠BMO=∠BMC.

Hence,

∠AMC=∠AMO+∠BMO+∠BMC=3∠BMC.

apex says the answer is true

in counting the number of heads in 4 coin flips, the probability that we get exactly one head is the probability that we get anyone of the following 4 outcomes: httt, thtt, ttht, or ttth. each has probability 1/16, so the probability to get exactly one head in 4 flips is 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4.