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Question

Let $$\overset{\leftrightarrow}{AB}$$ be a straight line. Let $$\overset{\leftrightarrow}{CD}$$ and $$\overset{\leftrightarrow}{EF}$$ be two straight lines such that each of them is perpendicular to $$\overset{\leftrightarrow}{AB}$$. Prove that $$\overset{\leftrightarrow}{CD}\parallel \overset{\leftrightarrow}{EF}$$.
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Solution

Let $$\overset{\leftrightarrow}{AB}$$ intersect $$\overset{\leftrightarrow}{CD}$$ and $$\overset{\leftrightarrow}{EF}$$ at L and M respectively.
Since $$\overset{\leftrightarrow}{CD}\bot \overset{\leftrightarrow}{AB}$$, 
we have $$\angle DLA =90^o$$. 
Using $$\overset{\leftrightarrow}{EF}\bot \overset{\leftrightarrow}{AB}$$, 
we also get $$\angle FMA = 90^o$$. 
Thus $$\angle DLA =\angle FMA$$.
But these are corresponding angles made by the transversal $$\overset{\leftrightarrow}{AB}$$ with the lines $$\overset{\leftrightarrow}{CD}$$ and $$\overset{\leftrightarrow}{EF}$$. 
Hence, we conclude that $$\overset{\leftrightarrow}{CD}\parallel \overset{\leftrightarrow}{EF}$$.

Mathematics

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