How to solve the problem of Triangles of exercise 6.3 of class 10th, I don’t know how to solve this question please guide me the best way to solve this problem. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD

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# D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD Q.13

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Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.

In ΔADC and ΔBAC,

∠ADC = ∠BAC (Already given)

∠ACD = ∠BCA (Common angles)

∴ ΔADC ~ ΔBAC (AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

∴ CA/CB = CD/CA

⇒ CA

^{2}= CB.CD.Hence, proved.