D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that

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Solution

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.

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Similar questions

∠ A D C = ∠ B A C

C A^{2} = C B . C D

D is a point on the side BC of a $Δ A B C$ such that $∠ A D C = ∠ B A C$ . Show that $C A^{2} = C B . C D$ .

D is a point on the side BC of $△$ ABC such that $∠ A D C = ∠ B A C$ . Prove that : $C A^{2} = C B \times C D$ .

∠ A D C = ∠ B A C

C A^{2}

D

B C

△ A B C

∠ A D C = ∠ B A C .

\frac{C A}{C D} = \frac{C B}{C A}

C A^{2} = C B \times C D

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