AB is the diameter of the circle centered at point O. On the segment OB, as on the diameter, a circle is constructed

| April 4, 2021 | education

AB is the diameter of the circle centered at point O. On the segment OB, as on the diameter, a circle is constructed with the center at point O1. The chord of the larger circle BC meets the smaller circle at point E. A straight line is drawn through the points O1 and E, which intersects the large circle at points K and F (K – E – F), KE = 2 cm, EF = 8 cm. Find BC.

Let us prove that triangles ABC and OBE are similar.

Both triangles are rectangular, since their central angles ACB and OEB are based on the diameters of the circles. Angle ABC = EВO, then right-angled triangles are similar in acute angle. The coefficient of similarity of triangles is 1/2, since OB is the radius of the larger circle, and AB is the diameter. Then CB / EB = 1 / 2. CB = 2 * EB.

Then CE = EB.

By the property of intersecting chords CB and KF CE * EB = KE * EF.

CE2 = 2 * 8 = 16.

CE = 4 cm.

BC = 2 * 4 = 8 cm.

Answer: The length of the BC chord is 8 cm.



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