CDS 2026 (I) Elementary Mathematics, question 48

Topic: Geometry

Answers by SolvedPaper. UPSC has not released the official answer key for this paper yet. We will update it when they do.

Q48

ABC is a triangle in which AB=ACAB=AC and D is any point on BC. Which of the following statements is/are correct ?

I. AB2−AD2=CD×BDAB^{2}-AD^{2}=CD\times BD

II. BC2+BD2−DC2=2BD×BCBC^{2}+BD^{2}-DC^{2}=2BD\times BC

Select the answer using the code given below :

  1. (a) I only
  2. (b) II only
  3. (c) Both I and IICorrect answer
  4. (d) Neither I nor II

Solution.

Answer: (c) Both I and II

  1. Let M be the midpoint of BC. Since AB=AC,AMAB = AC,AM is perpendicular to BC. Write BM=CM=tBM = CM = t and MD=dMD = d.
  2. Pythagoras gives AB2−AD2=(AM2+t2)−(AM2+d2)=t2−d2AB^{2} - AD^{2} = \left( AM^{2} + t^{2} \right) - \left( AM^{2} + d^{2} \right) = t^{2} - d^{2}.
  3. The lengths BD and DC are t+dt + d and t−dt - d, in either order, so BD×DC=t2−d2BD \times DC = t^{2} - d^{2}.
  4. Thus I is true. For II, put BD=uBD = u and DC=vDC = v, so BC=u+vBC = u + v.
  5. Then BC2+BD2−DC2=(u+v)2+u2−v2=2u(u+v)=2BD×BCBC^{2} + BD^{2} - DC^{2} = (u + v)^{2} + u^{2} - v^{2} = 2u(u + v) = 2BD \times BC.