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1Let \(E\) be a point on the side \(CD\) of a square \(ABCD\). The point \(F\) lies on the line \(AB\) but not on the segment \(AB\), and satisfies \(|BF| = |DE|\). Prove that the lines \(AC\) and \(EF\) …Open2In triangle \(ABC\) the side \(AB\) is twice as long as \(AC\), that is \(|AB| = 2|AC|\). The point \(D\) lies on the ray \(CA\) and satisfies \(|CD| = 3|AC|\). Prove that the perpendicular dropped from …3Let \(AB\) be a segment and let \(P\) be any point of it other than \(A\) and \(B\). On the hypotenuses \(AP\) and \(PB\) erect isosceles right triangles \(APQ\) and \(PBR\), with the right angles at the …4A circle \(k\) has diameter \(AB\). A point \(M\) of \(k\) is chosen, different from \(A\) and from \(B\). Let \(k_1\) be the circle with centre \(M\) that touches the diameter \(AB\). The line \(AB\) …5Let \(E\) and \(F\) be the midpoints of the sides \(AD\) and \(DC\) of a rectangle \(ABCD\), and let \(G\) be the point where the segments \(AF\) and \(EC\) cross. Prove that \[ \angle CGF = \angle FBE . \] …6A square \(ABCD\) is given, together with points \(E\) and \(F\) lying outside the square such that the triangles \(BEC\) and \(CFD\) are equilateral. Prove that the triangle \(AEF\) is equilateral as …7Let \(ABC\) be a triangle. Points \(D\) and \(E\) lie on the rays \(CA\) and \(CB\) respectively, but not on the sides of the triangle \(ABC\), and are chosen so that \[ |AD| = |BE| = |AB| . \] Let \(G\) …8A disc \(K\) of radius \(R\) is cut into three circular sectors in such a way that the areas of the two smaller sectors add up to the area of the largest sector, while the difference of the areas of the …9Three squares are inscribed in a right triangle \(ABC\) with the right angle at \(B\), arranged as in the picture: the largest one has two sides lying on the legs of the triangle, and each of the two smaller …10The triangle \(ABC\) has side lengths \(|AB| = 15\) cm, \(|BC| = 14\) cm and \(|CA| = 13\) cm. Let \(D\) be the foot of the altitude drawn from \(A\) to the side \(BC\), and let \(E\) be the point of that …11Let \(ABCD\) be a convex quadrilateral and let \(E\) and \(F\) be points on the sides \(AB\) and \(AD\) respectively, chosen so that \(EF \parallel BD\). The segment \(CE\) meets the diagonal \(BD\) at …12In a rectangle \(ABCD\) with \(|AB| > |BC|\), the perpendicular bisector of the diagonal \(AC\) meets the side \(CD\) at the point \(E\). The circle with center \(E\) and radius \(|AE|\) meets the side …13A triangle \(ABC\) has a point \(D\) on side \(AB\) and a point \(E\) on side \(AC\) such that \[ |AE| = |ED| = |DB| \qquad\text{and}\qquad |AD| = |DC| = |CB| . \] Determine the angles of triangle \(ABC\). …14A parallelogram \(ABCD\) satisfies \(|AB| = |BD|\). Let \(K\) be the point of line \(AB\), different from \(A\), with \(|KD| = |AD|\). Let \(M\) be the image of \(C\) under the half-turn about \(K\) (so …15The lines containing the two diagonals of a quadrilateral \(ABCD\) meet at an angle of \(60^{\circ}\). Each vertex of the quadrilateral is reflected in the line containing the diagonal that joins its two …16Natasa glued a square and an equilateral triangle of the same side length into a pentagon. Out of seven copies of that pentagon she built the figure shown on the right, which sits inside a large square. …17Three curves are drawn inside a square \(ABCD\): the quarter circle \(\mathcal{Q}\) centred at the vertex \(A\) and passing through \(B\) and \(D\); the semicircle \(\mathcal{P}\) centred at the midpoint …18Let \(M\) be the midpoint of the base \(AB\) of a trapezoid \(ABCD\), so that \(AB \parallel CD\). A point \(E\) is taken in the interior of the segment \(AC\) in such a way that the lines \(BC\) and \(ME\) …19Let \(I\) be the incenter of a triangle \(ABC\), and suppose that \[ |CA| + |AI| = |BC| . \] Determine the ratio of the sizes of the angles \(\angle BAC\) and \(\angle CBA\).20Let \(I\) be the centre of the inscribed circle of a triangle \(ABC\), and let \(A_1\), \(B_1\), \(C_1\) be the feet of the perpendiculars dropped from \(I\) to the sides \(BC\), \(AC\) and \(AB\). The …21Let \(D\) be the midpoint of side \(AB\) of an acute triangle \(ABC\). Points \(A'\) and \(B'\) are chosen on the segments \(AC\) and \(BC\) so that the triangles \(ADA'\) and \(DBB'\) are both isosceles …22In a hexagon \(ABCDEF\) the following hold: \(\angle BAF = 150^{\circ}\), \(\angle ACB = \angle ADC = 90^{\circ}\), \(|AC| = |BC|\), triangle \(ABC\) is similar to triangle \(ADE\), and triangle \(BCD\) …
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