In the figure, $ABCD$ is a trapezium in which $AB$ is parallel to $DC$ and \(\angle ADC = \ang

Question

In the figure, $ABCD$ is a trapezium in which $AB$ is parallel to $DC$ and $\angle ADC = \angle BCA$. Prove that $\Delta ABC$ is similar to $\Delta CAD$, $AC^2 = AB \times CD$.

Accepted Answer

(i) Proved: $\triangle ABC\sim\triangle CAD$ (ii) Proved: $AC^2=AB\times CD$

Problems involving Properties of Similar Triangles

Problems involving Properties of Similar Triangles

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