By V. Prasolov D.Leites (translator)
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Additional info for Problems in Plane and Solid Geometry v.1 Plane Geometry
Then by subtracting from their sum the equality ⌣ BA′ + ⌣ CA′ = 2∠A we get ⌣ C ′ B ′ =⌣ C ′ A+ ⌣ AB ′ = 480◦ − 2(∠A + ∠B + ∠C) = 120◦ . Similarly, ⌣ B ′ A′ =⌣ C ′ A′ = 120◦ . 19. a) Let us prove, for example, that AA1 ⊥ C1 B1 . Let M be the intersection point of these segments. Then ∠AM B1 = ⌣ AB1 + ⌣ A1 B+ ⌣ BC1 = ∠ABB1 + ∠A1 AB + ∠BCC1 = 2 ∠B + ∠A + ∠C = 90◦ . 2 b) Let M1 and M2 be the intersection points of segments AA1 with BC and BB1 with AC. Right triangles AM1 C and BM2 C have a common angle ∠C; hence, ∠B1 BC = ∠A1 AC.
Since these lines are parallel, AD1 : D1 O1 = O2 P1 : P1 O1 and D2 O2 : AD2 = O2 P2 : P2 O1 . The similarity of quadrilaterals AKO1 M and O2 N AL yields AD1 : D1 O1 = D2 O2 : AD2 . , P1 = P2 . CHAPTER 2. INSCRIBED ANGLES Background 1. Angle ∠ABC whose vertex lies on a circle and legs intersect this circle is called inscribed in the circle. Let O be the center of the circle. Then ∠ABC = 1 ∠AOC 2 ◦ 180 − 21 ∠AOC if points B and O lie on one side of AC otherwise. The most important and most often used corollary of this fact is that equal chords subtend angles that either are equal or the sum of the angles is equal to 180◦ .
62. Circle S1 with diameter AB intersects circle S2 centered at A at points C and D. Through point B a line is drawn; it intersects S2 at point M that lies inside S1 and it intersects S1 at point N . Prove that M N 2 = CN · N D. 63. Through the midpoint C of an arbitrary chord AB on a circle chords KL and M N are drawn so that points K and M lie on one side of AB. Segments KN and M L intersect AB at points Q and P , respectively. Prove that P C = QC. 64. a) A circle that passes through point C intersects sides BC and AC of triangle ABC at points A1 and B1 , respectively, and it intersects the circumscribed circle of triangle ABC at point M .
Problems in Plane and Solid Geometry v.1 Plane Geometry by V. Prasolov D.Leites (translator)