D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC prove that BC by CA is equal to CA/CD
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Answer:
In ΔADC and ΔBAC,
∠ADC = ∠BAC (Given)
∠ACD = ∠BCA (Common angle)
∴ ΔADC ~ ΔBAC (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
∴ CA/CB =CD/CA
⇒ CA2 = CB.CD.
An alternate method:
Given in ΔABC, ∠ADC = ∠BAC
Consider ΔBAC and ΔADC
∠ADC = ∠BAC (Given)
∠C = ∠C (Common angle)
∴ ΔBAC ~ ΔADC (AA similarity criterion)
⇒
∴ CB x CD = CA²
Hope This Helps :)
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