In the diagram, $AB$ is parallel to $DC$, and $AC$ and $BD$ meet at $X$. Prove that triang

Question

In the diagram, $AB$ is parallel to $DC$, and $AC$ and $BD$ meet at $X$. Prove that triangles $\Delta ABX$ and $\Delta CDX$ are similar. Given that $AB = 24\text{ cm}$, $AX = 12\text{ cm}$ and $XC = 18\text{ cm}$. Find $CD$. Given also that $BD = 42\text{ cm}$, find $BX$. Find $\frac{\text{Area}\hspace{0.5em} \Delta ABX}{\text{Area}\hspace{0.5em} \Delta CDX}$ , $\frac{\text{Area}\hspace{0.5em} \Delta ADX}{\text{Area}\hspace{0.5em} \Delta ACD}$ .

Accepted Answer

$\text{(i)}\ \triangle ABX\sim\triangle CDX\text{ (AA)}$ $\text{(ii)}\ CD=36\text{ cm}$ $\text{(iii)}\ BX=16.8\text{ cm}$ $\text{(iv)(a)}\ \dfrac{4}{9}$ $\text{(iv)(b)}\ \dfrac{2}{5}$

Problem involving Similar Figures

Problem involving Similar Figures

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