Question
$D$ is a point on side $B C$ of $\triangle A B C$ such that, $\angle A D C=\angle B A C$. Show that $A C^2=B C \times D C$.
✓
Answer
To Prove: $A C^2=B C \times D C$
Consider $\triangle A D C$ and $\triangle B A C$.
$\angle A D C=\angle B A C$(Given)
Also,
$\angle A C D=\angle B C A \quad$ (Common angle)
Therefore,
$\triangle A D C \sim \triangle B A C$ (by AA similarity test)
Hence, the corresponding sides of similar triangles are proportional.
$\frac{A C}{B C}=\frac{D C}{A C}$
On cross-multiplying, we get
$A C^2=B C \times D C$
Consider $\triangle A D C$ and $\triangle B A C$.
$\angle A D C=\angle B A C$(Given)
Also,
$\angle A C D=\angle B C A \quad$ (Common angle)
Therefore,
$\triangle A D C \sim \triangle B A C$ (by AA similarity test)
Hence, the corresponding sides of similar triangles are proportional.
$\frac{A C}{B C}=\frac{D C}{A C}$
On cross-multiplying, we get
$A C^2=B C \times D C$
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