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The previous case can be extended to cover the case where the measure of the inscribed angle is the difference between two inscribed angles as discussed in the first part of this proof. Given a circle whose center is point O, choose three points V, C, D on the circle. Draw lines VC and VD: angle ∠DVC is an inscribed angle.
Since OA = OB = OC, OBA and OBC are isosceles triangles, and by the equality of the base angles of an isosceles triangle, ∠ OBC = ∠ OCB and ∠ OBA = ∠ OAB. Let α = ∠ BAO and β = ∠ OBC. The three internal angles of the ∆ABC triangle are α, (α + β), and β. Since the sum of the angles of a triangle is equal to 180°, we have
In geometry, the incircle or inscribed circle of a triangle is the largest circle that can be contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is a triangle center called the triangle's incenter .
The vertices of every triangle fall on a circle called the circumcircle. (Because of this, some authors define "concyclic" only in the context of four or more points on a circle.) [2] Several other sets of points defined from a triangle are also concyclic, with different circles; see Nine-point circle [3] and Lester's theorem.
An inscribed angle (examples are the blue and green angles in the figure) is exactly half the corresponding central angle (red). Hence, all inscribed angles that subtend the same arc (pink) are equal. Angles inscribed on the arc (brown) are supplementary. In particular, every inscribed angle that subtends a diameter is a right angle (since the ...
Finally, the angles CME and FMA are the same. Hence, AFM is an isosceles triangle, and thus the sides AF and FM are equal. The proof that FD = FM goes similarly: the angles FDM, BCM, BME and DMF are all equal, so DFM is an isosceles triangle, so FD = FM. It follows that AF = FD, as the theorem claims.
The theorem follows directly from the fact that the triangles PAC and PBD are similar. They share ∠ DPC and ∠ ADB = ∠ ACB as they are inscribed angles over AB . The similarity yields an equation for ratios which is equivalent to the equation of the theorem given above: P A P C = P B P D ⇔ | P A | ⋅ | P D | = | P B | ⋅ | P C ...
Using the usual notations for a triangle (see the figure at the upper right), where a, b, c are the lengths of the three sides, A, B, C are the vertices opposite those three respective sides, α, β, γ are the corresponding angles at those vertices, s is the semiperimeter, that is, s = a + b + c / 2 , and r is the radius of the inscribed circle, the law of cotangents states that