D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC . SHOW that
CA²=CB.CD.
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Answer:
In △ADC and △BAC
∠ADC=∠BAC (Given)
∠C is Common
∴ by AA Criterion of Similarity, △ADC ∼ △BAC
⇒
BA
AD
=
AC
DC
=
BC
AC
⇒
AC
DC
=
BC
AC
∴ CA
2
=CB.CD
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