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

    Ncert solutions class 10 chapter 6-24

    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

    ⇒ CA2 = CB.CD.

    Hence, proved.

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