By Viktor Prasolov, Dimitry Leites (translator)

**Read Online or Download Problems in Plane and Solid Geometry, Volume 1, Plane Geometry PDF**

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**Extra info for Problems in Plane and Solid Geometry, Volume 1, Plane Geometry **

**Example text**

62. Let D be the midpoint of segment BH. Since △BHA ∼ △HEA, it follows that AD : AO = AB : AH and ∠DAH = ∠OAE. Hence, ∠DAO = ∠BAH and, therefore, △DAO ∼ △BAH and ∠DOA = ∠BAH = 90◦ . 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 .

Given triangle ABC; on its side AB point P is chosen; lines P M and P N parallel to AC and BC, respectively, are drawn through P so that points M and N lie on sides BC and AC, respectively; let Q be the intersection point of the circumscribed circles of triangles AP N and BP M . Prove that all lines P Q pass through a fixed point. 69. The continuation of bisector AD of acute triangle ABC inersects the circumscribed circle at point E. Perpendiculars DP and DQ are dropped on sides AB and AC from point D.

Since ∠EBA = 90◦ , the distance from point O to AB is equal to 12 EB. 76. Let the perpendicular dropped from point P to BC intersect BC at point H and AD at point M (Fig. 16). SOLUTIONS 51 Figure 16 (Sol. 76) Therefore, ∠BDA = ∠BCA = ∠BP H = ∠M P D. Since angles M DP and M P D are equal, M P is a median of right triangle AP D. , AM = P M = M D. 77. The midpoints of the sides of quadrilateral ABCD are vertices of a rectangle (cf. 2), hence, they lie on one circle. Let K and L be the midpoints of sides AB and CD, let M be the intersection point of lines KP and CD.