Question

D is a point on the side BC of a triangle ABC such that angleADC equal angleBAC Show that CA2equal CB. CD​

Answer 1

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)

⇒AB/AD = CB/CA= CA/CD

Consider,

                 CB/CA = CA/CD

                  ∴ CB x CD = CA²

Answer 2

Answer:

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