Question

D is a point on side BC of a triangle ABC such that angle ADC is equal to angle BAC prove that CD is equals to C B divided by CA

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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

Answer 2

Step-by-step explanation:

Given that D is point on side BC of Triangle ABC, such that Angle ADC = Angle BAC

To prove CD = CB/CA

Proof:

It's given that Angle BAC = Angle ADC

Also Angle C = Angle C (common)

Therefore, by AA Similarity, Triangle BCA ~ Triangle ACD

=> CB/CA = CA/CD = AB/AD

or CB/CA = CA/CD

or CD = CA^2/CB.... Answer