Diameter KF and chord KP / are drawn in a circle centered at point O. A tangent line is drawn through point P

| August 18, 2021 | education

Diameter KF and chord KP / are drawn in a circle centered at point O. A tangent line is drawn through point P, which intersects the KF beam at point E at an angle of 30 degrees. Prove that triangle KPE is isosceles.

In triangle OPE: angle P = 90 ° (tangent and radius are perpendicular).

OE = 2 * PO = 2R (opposite an angle of 30 ° lies a leg, half the size of the hypotenuse).

Let us express the cosine of the angle PEO: cosPEO = PE / OE; cos30 ° = PE / 2R;

PE = 2R * cos30 ° = 2R * √3 / 2 = √3R.

In the POE triangle, the angle POE = 180 ° – (90 ° + 30 °) = 60 ° (the sum of the angles in the triangle is 180 °).

This means that the angle PОК = 180 ° – 60 ° = 120 ° (the angle POK and the angle POE are adjacent angles).

Triangle PОК – isosceles (PО = KO = R): angle К = angle Р = (180 ° – 120 °): 2 = 30 °.

Let us express the cosine of the angle P: cosP = KP / 2R; KP = 2R * cos30 ° = 2R * √3 / 2 = √3R.

Hence, KP = PE = √3R and, therefore, the KRE triangle is isosceles.



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