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$D M$ is a diameter of circle $A . R T D \perp R C$, $R T=3 \mathrm{~cm}, R C=4 \mathrm{~cm}, M C=6 \mathrm{~cm}$ and $D M=2 x$.

1. Prove $\triangle C R T \| \triangle C D M$
2. Calculate $x$

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\begin{aligned} &\text { 1. } \hat{C}_{3}=90^{\circ}(\angle \text { in semi-circle) } \\ &\text { In } \triangle C R T \text { and } \triangle C D M \\ &\hat{R}_{2}=\hat{C}_{3}=90^{\circ} \text { (given and proven) } \\ &\hat{T}_{1}=\hat{M} \text { (ext. } \angle \text { of cyc. quad) } \\ &\left.\hat{C}_{1}=\hat{D}_{2} \text { (sum } \angle \text { 's of } \triangle\right) \\ &\therefore \triangle R T C \| \triangle C M D \text { (AAA) } \\ &\therefore \frac{R T}{C M}=\frac{T C}{M D}=\frac{R C}{C D}(\| \triangle \text { 's) } \end{aligned}
\begin{aligned} &\text { 2. } T C^{2}=3^{2}+4^{2} \text { (Pythag.) } \\ &\therefore T C=5 \mathrm{~cm} \\ &\frac{R T}{C M}=\frac{T C}{M D} \\ &\frac{3}{6}=\frac{5}{2 x} \\ &6 x=30 \\ &\therefore x=5 \mathrm{~cm} \end{aligned}
by Diamond (58,513 points)

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