D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BA

1.	D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.
Answer» Data: D is a point on the side BC of a triangle ∆ABC such that ∠ADC = ∠BAC. To Prove: CA² = CB × CD Let ∠ADC = ∠BAC = 100° In ∆ABC, If ∠B = 50°,then ∠C = 30° In ∆ADC, If ∠C = 30°. then ∠DAC = 50° In ∆BCP, ∠A= 100°, ∠B= 50°, ∠C= 30° In ∆ADC, ∠ADC = 100. ∠DAC = 50°. ∠ACD = 30° Similarity criterion of ∆ is A.A.A. ∴ In ∆ABC and ∆ADC, \(\frac{CA}{BC} = \frac{DC}{CA}\) ∴ CA × CA = BC × DC ∴ CA²= BC × DC.

D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.

Answer»

Data: D is a point on the side BC of a triangle ∆ABC such that ∠ADC = ∠BAC.

To Prove: CA² = CB × CD

Let ∠ADC = ∠BAC = 100°

In ∆ABC, If ∠B = 50°,then ∠C = 30°

In ∆ADC, If ∠C = 30°. then ∠DAC = 50°

In ∆BCP, ∠A= 100°, ∠B= 50°, ∠C= 30°

In ∆ADC, ∠ADC = 100. ∠DAC = 50°. ∠ACD = 30°

Similarity criterion of ∆ is A.A.A.

∴ In ∆ABC and ∆ADC,

\(\frac{CA}{BC} = \frac{DC}{CA}\)

∴ CA × CA = BC × DC

∴ CA²= BC × DC.