By Viktor Prasolov, Dimitry Leites (translator)

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Moreover, AK BL and AL ⊥ BK. It follows that AKLB is a rhombus. 74. In the circumscribed circle take a point D′ so that DD′ AC. Since DD′ ⊥ BD, it follows that BD′ is a diameter and, therefore, ∠D′ AB = ∠D′ CB = 90◦ . Hence, 1 1 SABCD = SABCD = (AD′ · AB + BC · CD′ ) = (AB · CD + BC · AD). 75. Let us draw diameter AE. Since ∠BEA = ∠BCP and ∠ABE = ∠BP C = 90◦ , it follows that ∠EAB = ∠CBP . The angles that intersect chords EB and CD are equal, hence, EB = CD. Since ∠EBA = 90◦ , the distance from point O to AB is equal to 12 EB.

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.

Let M be the midpoint of AC, N the midpoint of BD. We have AM 2 = AO2 − OM 2 and BN 2 = BO2 − ON 2 ; hence, AC 2 + BD2 = 4(R2 − OM 2 ) + 4(R2 − ON 2 ) = 8R2 − 4(OM 2 + ON 2 ) = 8R2 − 4OP 2 since OM 2 + ON 2 = OP 2 . 73. The correspondiong legs of acute angles ∠BLP and ∠BDC are perpendicular, hence, the angles are equal. Therefore, ∠BLP = ∠BDC = ∠BAP . Moreover, AK BL and AL ⊥ BK. It follows that AKLB is a rhombus. 74. In the circumscribed circle take a point D′ so that DD′ AC. Since DD′ ⊥ BD, it follows that BD′ is a diameter and, therefore, ∠D′ AB = ∠D′ CB = 90◦ .

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Problems in Plane and Solid Geometry, Volume 1, Plane Geometry by Viktor Prasolov, Dimitry Leites (translator)


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