Question

in fig. in triangle ABC, point D on side BC is such that angle bac​

Answer 1

Answer:

Step-by-step explanation:

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 :)