Question

D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC Show that CA square=CB

Answer 1

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)

⇒ $\frac{AB}{AD} = \frac{CB}{CA} = \frac{CA}{CD}$

$Consider, \frac{CB}{CA} = \frac{CA}{CD}$

∴ CB x CD = CA²

Hope This Helps :)

Answer 2

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