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