Posts

practice problems pending

Image
 Q1. ABC is a triangle and D and E are interior points of the sides AB and BC respectively such that: AD/DB = 1/3 CE/EB = 3 If AE and CD intersect at F, find CF/FD. S1. Approach 1: Using mass points : AD/DB = 1/3 => A has higher mass since it's closer to D. Let mA = 3 mB = 1 => mC = 1/3 => mE = 4/3 Now mF = 3 + 4/3 = 13/3 and it matches 4 + 1/3 = 13/3 So we have assigned masses correctly. => CF/FD = 4/(1/3) = 12 Approach 2: Menelaus theorem Typically you would notice that if we have a solution using mass points, we can also solve it using Ceva's theorem or Menelaus' theorem. Here, in triangle BDC, AFE is the traversal intersecting all sides. BE/EC * CF/FD * DA/AB = 1 => 1/3 * CF/FD * 1/4 = 1 => CF/FD = 12 = Answer Q2.  L and M are the mid-points of the diagonals BD and AC respectively of the quadrilateral ABCD. Through D, draw DE equal and parallel to AB. Show that EC || LM EC = 2LM S2. It's straightforward with co-ordinate geometry. A (0,0) B(x1,0) C(...

practice problems pending

Image
 Q1. In quadrilateral ABCD, the diagonals AC and BD meet at O. Suppose the four triangles AOB,BOC,COD and DOA are equal in area, prove that ABCD is a parallelogram. S1. [AOB] = [COD] => 1/2.OA.OB.Sin(AOB) = 1/2.OC.OD.Sin(COD) Angles AOB and COD are same. => OA.OB = OC.OD Similarly, OA.OD = OC.OB Multiply both => OA = OC and OB = OD => O bisects both diagonals. If the diagonals bisect each other, the quadrilateral is a ||gram. Why? Let the quadrilateral be A (0,0), B(x,0), C(a,b) D(p,q) O = Midpoint of AC = (a/2,b/2) = Midpoint of BD = ((p+x)/2,q/2) b/2 = q/2 => b = q => CD || AB and a = p+x Slope of AD = q/p Slope of BC = b/(a-x) = q/p => AD || BC Opposite sides are || => ABCD is a ||gram. H.P. Q2. In a parallelogram  A B C D A BC D , a point  P P on the segment  A B A B is taken such that  A P A B = A B A P ​ =   61 2022 2022 61 ​  and a point  Q Q on the segment  A D A D is taken such that  A Q A D = 61 2065...