The median BM of triangle ABC is the diameter of the circle intersecting side BC in its middle. The length of the side AC is 4. Find the radius of the circumscribed circle of triangle ABC
Draw a segment MP, as shown in the figure. BM is the diameter of the small circle (according to the condition of the problem), therefore the triangle BMP is right-angled with the hypotenuse BM (by the property of the circumscribed circle).
Consider the triangles BMP and CPM:
MP – common side
BP = PC (by condition of the problem)
∠BPM = ∠CPM, because ∠BPM is direct and ∠CPM is adjacent to it.
Therefore, the triangles BMP and CPM are equal (by the first sign). It follows that BM = MC = MA.
Consider the triangle BMC. Because MB = MC, then this triangle is isosceles, therefore ∠MCP = ∠PBM (by the property of isosceles triangles).
The triangle ABM has a similar situation, ∠BAM = ∠ABM. Those. it turns out that ∠BAM + ∠MCP = ∠ABC. From the theorem on the sum of the angles of a triangle it follows that 180 ° = ∠BAM + ∠MCP + ∠ABC
180 ° = ∠ABC + ∠ABC
180 ° = 2 * ∠ABC
90 ° = ∠ABC
It follows that the triangle ABC is rectangular. By the property of the circumscribed circle, it follows that the point M is the center of the circle => R = AC / 2 = 4/2 = 2.
Answer: R = 2.
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