By Prasolov V.V.
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Extra resources for Problems in plane and solid geometry, V.1, Plane geometry
Example text
53. Points B1 and H lie on the circle with diameter AB, hence, ∠(AB, BC) = ∠(AB, BH) = ∠(AB1 , B1 H) = ∠(B1 C1 , B1 H). Similarly, ∠(AC, BC) = ∠(B1 C1 , C1 H). 54. On an extension of segment BP beyond point P take point D such that P D = CP . Then triangle CDP is an equilateral one and CD QP . , P1Q = CP + BP . 55. Segment QE subtends angles of 45◦ with vertices at points A and B, hence, quadrilateral ABEQ is an inscribed one. Since ∠ABE = 90◦ , it √ follows that ∠AQE = 90◦ . √ AF AE Therefore, triangle AQE is an isosceles right triangle and AQ = 2.
63. Let AA1 , BB1 and CC1 be heights of triangle ABC. Let us drop from point B1 perpendiculars B1 K and B1 N to sides AB and BC, respectively, and perpendiculars B1 L and B1 M to heights AA1 and CC1 , respectively. Since KB1 : C1 C = AB1 : AC = LB1 : A1 C, it follows that △KLB1 ∼ △C1 A1 C and, therefore, KL C1 A1 . Similarly, M N C1 A1 . Moreover, KN C1 A1 (cf. 53). It follows that points K, L, M and N lie on one line. 64. a) Let O be the midpoint of AC, let O1 be the midpoint of AB and O2 the midpoint of BC.
1. 1. Vertex A of an acute triangle ABC is connected by a segment with the center O of the circumscribed circle. From vertex A height AH is drawn. Prove that ∠BAH = ∠OAC. 2. Two circles intersect at points M and K. Lines AB and CD are drawn through M and K, respectively; they intersect the first circle at points A and C, the second circle at points B and D, respectively. Prove that AC BD. 3. From an arbitrary point M inside a given angle with vertex A perpendiculars M P and M Q are dropped to the sides of the angle.