By V. Prasolov D.Leites (translator)
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Additional resources for Problems in Plane and Solid Geometry v.1 Plane Geometry
For the other projections the proof is similar. 78. a) It is worth to observe that since points A, B, C and D divide the circle into arcs smaller than 180◦ each, then the quadrilateral constructed contains this circle. The angle ϕ between the tangents drawn through points A and B is equal to 180◦ − ∠AOB and the angle ψ between the tangents drawn through points C and D is equal to 180◦ − ∠COD. Since ∠AOB + ∠COD = 180◦ , it follows that ϕ + ψ = 180◦ . Remark. , AC ⊥ BD. b) Let O be the center of the inscribed circle.
The extensions of sides AB and CD of the inscribed quadrilateral ABCD meet at point P ; the extensions of sides BC and AD meet at point Q. Prove that the intersection points of the bisectors of angles ∠AQB and ∠BP C with the sides of the quadrilateral are vertices of a rhombus. §6. 41. The inscribed circle of triangle ABC is tangent to sides AB and AC at points M and N , respectively. Let P be the intersection point of line M N with the bisector (or its extension) of angle ∠B. Prove that: a) ∠BP C = 90◦ ; b) SABP : SABC = 1 : 2.
Let K and L be the midpoints of sides AB and CD, let M be the intersection point of lines KP and CD. 76 P M ⊥ CD; hence, M is the projection of point P on side CD and point M lies on the circle with diameter KL. For the other projections the proof is similar. 78. a) It is worth to observe that since points A, B, C and D divide the circle into arcs smaller than 180◦ each, then the quadrilateral constructed contains this circle. The angle ϕ between the tangents drawn through points A and B is equal to 180◦ − ∠AOB and the angle ψ between the tangents drawn through points C and D is equal to 180◦ − ∠COD.