In ΔACD, B and E are two points on side AC and AD respectively, such that BE is parallel to CD. CD=9 cm. BE = 6 cm, AB = 5 cm and ED=2cm. What are the measures of the lengths (in cm) of AE and BC?

4, 2.5

Here,BE is parallel to CD

So, \(\frac{AB}{AC}\) = \(\frac{BE}{CD}\) = \(\frac{AE}{AD}\)

Then, \(\frac{AB}{AC}\) = \(\frac{6}{9}\)

⇒ \(\frac{5}{5\; + \;BC}\) = \(\frac{6}{9}\)
⇒ 6 BC + 30 = 45

⇒ 6BC = 45 - 30 = 15

⇒ BC = \(\frac{15}{6}\) = 2.5 cm

⇒ Also, \(\frac{AE}{AD}\) = \(\frac{6}{9}\) 

⇒ \(\frac{AE}{AE\;+\; 2}\) = \(\frac{6}{9}\) 

⇒ 9AE = 6AE + 12

⇒ 9AE - 6AE = 12

⇒ 3AE = 12

⇒ AE = \(\frac{12}{3}\) = 4 cm

Therefore, length of AE and BC are 4cm and 2.5 cm respectively.