(a) In the given figure, D is a point on the side BC of △ABC such that ∠ADC = ∠BAC. Show that CA²

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D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC | Exercise 6.3 class 10 Q13 D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC | Exercise 6.3 class 10 Q13
In the given figure, D is a point on the side BC of △ABC such that ∠ADC=∠BAC. Show that CA² = CD. CB In the given figure, D is a point on the side BC of △ABC such that ∠ADC=∠BAC. Show that CA² = CD. CB
In the given figure, D is a point on the side BC of △ABC such that ∠ADC= ∠BAC. Show that CA²= CD.CB. In the given figure, D is a point on the side BC of △ABC such that ∠ADC= ∠BAC. Show that CA²= CD.CB.
D is a point on the side BC of triangle ABC such that angle ACD = angle BAC show that AC^2= BC.DC . D is a point on the side BC of triangle ABC such that angle ACD = angle BAC show that AC^2= BC.DC .
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D is a point on the side BC of triangleABC such that angleADC= angleBAC, prove that CA^(2)=CBxxC... D is a point on the side BC of triangleABC such that angleADC= angleBAC, prove that CA^(2)=CBxxC...
In the given figure, D is a point on the side BC Of ∆ABC such that ∠ADC = ∠BAC Show that CA^2=CD .CB In the given figure, D is a point on the side BC Of ∆ABC such that ∠ADC = ∠BAC Show that CA^2=CD .CB
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. Show that CA2 = CB.CD D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. Show that CA2 = CB.CD