Question

D is a point on the side BC of a triangle ABC such that angle ADC=angke BAC.show that CA2=CB.CD.​

Answer 1

Answer:

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Answer 2

Answer:

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Step-by-step explanation:

I think it's your 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.

We know that corresponding sides of similar triangles are in proportion.∴ CA/CB =CD/CA

We know that corresponding sides of similar triangles are in proportion.∴ CA/CB =CD/CA⇒ CA2 = CB.CD. 

We know that corresponding sides of similar triangles are in proportion.∴ CA/CB =CD/CA⇒ CA2 = CB.CD. An alternate method:

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

∴ CB x CD = CA²Hope This Helps :)