Triangles and Polygons | Problems
- ABC is a triangle having BC = 2 AB. Bisect BC in D and BD in E. Prove that AD bisects ∠CAE.
- ABC is a triangle. D and E are any two points on AB and AC. The bisectors of the angles ABE and ACD meet in F. Show that ∠BDC – ∠BEC = 2 ∠BFC.
- ABC is a right-angled triangle at A and AB > AC. Bisect BC in D draw DE perpendicular to the hypotenuse BC to meet the bisector of the right angle A in E. Prove that (i) AD = DE; (ii) ∠DAE = ½ ( ∠C – ∠B).
- The sides AB, BC, and AC of a triangle are bisected in F, G and H respectively. If BE is drawn perpendicular to AC, prove that ∠FEG = ∠FHG (Fig.)
- In an isosceles triangle ABC, the vertex angle A is 2/7 right angle. Let D be any point on the base BC and take a point E on CA so that CE = CD. Join ED and produce it to meet AB produced in F (Fig.). If the bisector of the ∠EDC meets AC in R, show that (i) AE = EF; (ii) DR = DC.
- The vertex C is a right angle in the triangle ABC. If the points D and E are taken on the hypotenuse, so that BC = BD and AC = AE, show that DE equals the sum of the perpendiculars from D and E on AC and BC respectively.
- The side AB in the rectangle ABCD is twice the side BC. A point P is taken on the side AB so that BP = ¼ AB. Show that BD is perpendicular to CP.
- ABC is a triangle. BF, CG are any two lines drawn from the extremities of the base BC to meet AC and AB in F and G respectively and intersect in H. Show that AF + AG > HF + HG.
- ABCD is a square. The bisector of the ∠DBA meets the diagonal AC in F. If CK is drawn perpendicular to BF, intersecting BD in L and produced to meet AB in R, prove that AR = 2 SL, where S is the intersection
of the diagonals.
- The side AB in the triangle ABC is greater than AC, and D is the mid-point of BC. From C draw two perpendiculars to the bisectors of the internal and external vertical angles at A to meet them in F and G respectively. Prove that (i) DF =½( AB – AC ) ; (ii ) DG = ½( AB + AC ).
- ABC is a triangle and AD is any line drawn from A to the base BC. From B and C, two perpendiculars BE and CF are drawn to AD or AD produced. If R is the mid-point of BC, prove that RE = RF.
- Any finite displacement of a segment AB can be considered as though a rotation about a point called the pole. If the length of the segment and the angle which it rotates are given, describe the method to determine the pole. What would be the conditions of the angles between the rays around the pole?
- ABC is a triangle in which the vertical angle C is 60°. If the bisectors of the base angles A, B meet BC, AC in P, Q respectively, show that AB = AQ + BP.
- In any quadrilateral the two lines joining the mid-points of each pair of opposite sides meet at one point, with the line joining the mid-points of its diagonals.
- ABC is a triangle. If the bisectors of the two exterior angles B and C of the triangle meet at D and DE is the perpendicular from D on AB produced, prove that AE = ½ the perimeter of the ∆ABC.
- AB is a straight line. D and E are any two points on the same side of AB. Find a point F on AB so that (i) the sum of FD and FE is a minimum; (ii) the difference between FD and FE is a minimum. When is this impossible?
- Any straight line is drawn from A in the parallelogram ABCD. BE, CF, DG are perpendiculars from the other vertices to this line. Show that if the line lies outside the parallelogram, then CF = BE + DG, and when the line cuts the parallelogram, then CF = the difference between BE and DG.
- ABC is a triangle. AD and AE are two perpendiculars drawn from A on the bisectors of the base angles of the triangle B and C respectively. Prove that DE is parallel to BC.
- P and Q are two points on either side of the parallel lines AB and CD so that AB lies between P and CD. Two points L and M are taken on AB and CD. Find another two points X and Y on AB and CD so that ( PX + XY + YQ) is a minimum and XY is parallel to LM.
- Show that the sum of the two perpendiculars drawn from any point in the base of an isosceles triangle on both sides is constant.
- Draw a straight line parallel to the base BC of the triangle ABC and meeting AB and AC ( or produced ) in D, E so that DE will be equal to (i) the sum of BD and CE; (ii) their difference.
- ABCD is a parallelogram. From D a perpendicular DR is drawn to AC. BN is drawn parallel to AC to meet DR produced in N. Join AN to intersect BC in P. If DRN cuts BC in Q, prove that (i) P is the mid-point of BQ; (ii) AR = BN + RC.
- ABC is an isosceles triangle in which the vertical angle A = 120°. If the base BC is trisected in D and E, prove that ADE is an equilateral triangle.
- ABC and CBD are two angles each equal to 60°. A point O is taken inside the angle ABC and the perpendiculars OP, OQ, and OR are drawn from O to BA, BC, and BD respectively. Show that OR = OP + OQ.
- Construct a right-angled triangle, given the hypotenuse and the difference between the base angles.
- If the three medians of a triangle are known, construct the triangle.
- ABC is a triangle. On AB and AC as sides, two squares ABDE and ACFG are drawn outside the triangle. Show that CD, BF, and the perpendicular from A on BC meet in one point.
- If any point D is taken on the base BC of an isosceles triangle ABC and DEF is drawn perpendicular to the base BC and meets AB and AC or produced in E and F, show that ( DE + DF ) is a constant quantity and equals twice the perpendicular from A to BC.
- Any line is drawn through O, the point of concurrence of the medians of a triangle ABC. From A, B, and C three perpendiculars AP, BQ, and CR are drawn to this line. Show that AP = BQ + CR.
- Construct an isosceles trapezoid having given the lengths of its two parallel sides and a diagonal.
- On the sides of a triangle ABC, squares ABDE, ACFG, BCJK are constructed externally to it. BF, AJ, CD, AK are joined. If FC, DB are produced to meet AJ, AK in H, I respectively and X, Y are the intersections of BF, CD with AC, AB respectively, show that X, H are equidistant from C and Y, I are also equidistant from B. Prove also that the perpendiculars from A, B, C on GE, DK, FJ respectively intersect in the centroid of the triangle ABC and that GE, DK, FJ are respectively double the medians from A, B,C.
- The point of concurrence S of the perpendiculars drawn from the middle points of the sides of a triangle ABC, the orthocenter 0 , and the centroid G are collinear and OG = 2 SG.
* Problems are highly complicated and require high concentration and brain power with knowledge of Geometry. Try at your own risk.
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