In ΔABC, D is a point on side BC such that ∠ADC= ∠BAC. If CA = 12 cm, CD = 8 cm, then CB (in cm) = ?

18

Considering \(\Delta \)ACB and \(\Delta \)CDA

\(\angle\)ADC = \(\angle\)BAC (given)

\(\angle\)ACD = \(\angle\)ACB (common)

Thus, by AA postulate

\(\Delta \)CAB is similar to \(\Delta \)CDA

⇒ \(\frac{CA}{CB}\) = \(\frac{CD}{CA}\)

⇒ \(\frac{12}{CB}\) = \(\frac{8}{12}\)

⇒ CB = \(\frac{12\; × \;12}{8}\) = 18 cm.

Therefore, CB is 18 cm.