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

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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. D is a point on the side BC of a triangle ABC such that ∠ADC=∠BAC. Show that CA²=CB.CD.
(a) In the given figure, D is a point on the side BC of △ABC such that ∠ADC = ∠BAC. Show that CA² (a) In the given figure, D is a point on the side BC of △ABC such that ∠ADC = ∠BAC. Show that CA²
D is a point on the side BC of a triangle ABC such that angle ADC=angle BAC. Show that CA²=CB.CD. D is a point on the side BC of a triangle ABC such that angle ADC=angle BAC. Show that CA²=CB.CD.
CBSE Class-10 Maths NCERT solution -Triangles- Exercise 6.3 - Problem 13 CBSE Class-10 Maths NCERT solution -Triangles- Exercise 6.3 - Problem 13
E is a point on the side AD produced of a parallelogram ABCD | Q8 Ex 6.3 class 10 E is a point on the side AD produced of a parallelogram ABCD | Q8 Ex 6.3 class 10
In the figure, in triangle ABC, point D on side BC is such that, angle BAC = angle ADC. Prove that In the figure, in triangle ABC, point D on side BC is such that, angle BAC = angle ADC. Prove that
Similar Triangles Geometry - Class 10 Q.No.13 Exercise 6.3 NCERT Similar Triangles Geometry - Class 10 Q.No.13 Exercise 6.3 NCERT
Class 10 Ex 6.3 Q 13 | Chapter 6 | Triangles | NCERT Solutions | CBSE | EXERCISE 6.3 Class 10 Ex 6.3 Q 13 | Chapter 6 | Triangles | NCERT Solutions | CBSE | EXERCISE 6.3