By V. Prasolov D.Leites (translator)

**Read Online or Download Problems in Plane and Solid Geometry v.1 Plane Geometry PDF**

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**Additional resources for Problems in Plane and Solid Geometry v.1 Plane Geometry**

**Example text**

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.