Bunuel
Solution
Step 1: Analyse Question Stem
• ABC is a triangle and D lies on side AC.
• We need to find if area of triangle ABD is greater than the area of triangle BDC.
• Based on the above information let’s draw the below diagram:
o In this diagram, BE is the perpendicular drawn from B to AC

• Now, area of triangle \(ABD = \frac{1}{2}* base* height = \frac{1}{2}*AD*BE \)
• Also, area of triangle \(BDC = \frac{1}{2}*base*height = \frac{1}{2}*CD*BE\)
• we need to find if area of triangle ABD \(>\) area of triangle BDC
o Or, \(\frac{1}{2}*AD*BE > \frac{1}{2}*CD*BE ⟹ AD > CD\)
Therefore, we need to find out if \(AD > CD.\)
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: DC > AD
• According to this statement: \(AD < CD\)
•Thus, we can infer that area of triangle ABD \(<\) area of triangle BDC
Hence, statement 1 is sufficient and we can eliminate answer Options B, C and E.
Statement 2: BC > AB
• From this statement, we can’t find anything about AD and CD.
• So, let’s consider the
o area of triangle \(ABD = \frac{1}{2}* base* height = \frac{1}{2}*AB*GD\)
o and, area of triangle \(BDC = \frac{1}{2}*base*height = \frac{1}{2}*BC*DH\)
• We know that \(BC > AB\), however we don’t which one is greater between DG and DH.
• Thus, even after considering this case too, we can’t decidedly say whether area of triangle ABD is greater than area of triangle BDC or not.

Hence, statement 2 is not sufficient.
Thus, the correct answer is
Option A.