In the figure, in triangle ABC, point D on side BC is such that, angle BAC = angle ADC. Prove that

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In the figure in `DeltaABC,` point D on side BC is such that `/_BAC=/_ADC`. Prove that `CA^(2)= In the figure in `DeltaABC,` point D on side BC is such that `/_BAC=/_ADC`. Prove that `CA^(2)=
In ABC point D on side BC is such that DC=6, BC=15. Find A(∆ABD):A(∆ABC) and A(∆ABD):A(∆ADC). | 10th In ABC point D on side BC is such that DC=6, BC=15. Find A(∆ABD):A(∆ABC) and A(∆ABD):A(∆ADC). | 10th
In the given figure , D is a point on the side BC of `Delta ABC` such that `angle ADC=angle BAC`. . In the given figure , D is a point on the side BC of `Delta ABC` such that `angle ADC=angle BAC`. .
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 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...
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 .
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.
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.