In ΔABC point D on side BC is such that DC=6, BC=15.
Find A(ΔABD): A(ΔABC) and A(ΔABD) : A(ΔADC)
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Answer:
Step-by-step explanation:
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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