Step 1: According to the given measurements construct an imaginary figure.

              We have,

          AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 m


                      In ΔAPQ and ΔABC

Step 2: Make sure that the triangles (APQ and ABC) are similar.

             \angle A = \angle A (common angle)

             \frac{AP}{AB} = \frac{1}{3+1} = \frac{1}{4}   

             \frac{AQ}{AC} = \frac{1.5}{1.5 + 4.5} = \frac{1}{4}   

          So, \frac{AP}{AB} = \frac{AQ}{AC}

        since the sides are in proportion then the line PQ is parallel to BC

          Then \triangle APQ,\ \triangle ABC are similar (By SAS similarity)

Step 3: To prove the required equation, apply the Area of similar triangle theorem

       \frac{Area \triangle APQ}{Area \triangle ABC} = (\frac{AP}{AB})^2

       \frac{Area \triangle APQ}{Area \triangle ABC} = (\frac{1}{4})^2

       Area \triangle APQ = (\frac{1}{16}) * Area \triangle ABC

        Hence proved