In ΔABC, D is a point on side BC such that ∠ADC = ∠BAC. If CA = 15cm and CD = 9 cm, then CB (in cm) = ?

15 25 12  10

25

We have, CA = 15 cm and CD = 9 cm In ΔCAB and ΔCDA, ∠BAC = ∠ADC (Given) ∠ACB = ∠ACD (Common) So, ΔCAB ∼ ΔCDA   (by AA similarity) Then, \(\frac{CA}{CB}\) = \(\frac{CD}{CA}\) = \(\frac{15}{CB}\) = \(\frac{9}{15}\) = CB = (15 × 15)× \(\frac{1}{9}\)= 25 cm