In the given figure , D is a point on the side BC of `Delta ABC` such that `angle ADC=angle BAC`. Prove that `CA^(2)=CBxxCD`.

D is a point on the side BC of triangleABC such that angleADC= angleBAC , prove that CA^(2)=CBxxCD .

D is a point on the side BC of a DeltaABC such that AD bisects angle BAC . Then

D is the point on the side BC of ABC such that angleADC = angleBAC . Prove that, (BC)/(CA) = (CA)/(CD) .

D is a point on the side BC of a triangle ABC such that /_ADC= /_BAC . Show that CA^(2)= CB*CD .

D is a point on the side BC of ABC such that /_ADC=/_BAC. Prove that (CA)/(CD)=(CB)/(CA) or,CA^(2)=CB xx CD

In the given figure, D is a point on side BC of DeltaABC such that (BD)/(CD)=(AB)/(AC) . Prove that AD is the bisector of /_BAC .

D is a point on the side BC of DeltaABC . If /_ADC=/_BAC , then prove that CA^(2)=CBxxCD .

In the figure in DeltaABC, point D on side BC is such that /_BAC=/_ADC . Prove that CA^(2)=CBxxCD .

In the figure , in DeltaABC , point D on side BC is such that , DeltaBAC~=DeltaADC then prove that , CA^(2)-=CBxxCD .

D is a point on the side BC of triangle ABC such that angle ADC = angle BAC . If CA = 12 cm and BC = 16 cm, then the length of CD is :