Re: The length of segment
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17 Sep 2025, 06:37
Although this is not drawn to scale, we are given some hints as to the relationships of each point. As $D$ is the midpoint and $B D A$ and CDA both have $90^{\circ}$ angles, we know that there are two equal right angle triangles here which make up one isosceles (two equal sides) triangle. Given that $B D A$ and $C D A$ are mirror images of each other, $A C=A B$, and we can compare Column $A$ and Column $B$ by comparing $A E$ to $A B$.
Using the Pythagorean theorem $\(\left(a^2+b^2=\right. c^2\)$ ) we know:
$$
\(\begin{aligned}
& A D^2+B D^2=A B^2 \\
& A D^2+E D^2=A E^2
\end{aligned}\)
$$
$\(A D^2\)$ is common to both equations, so we can rearrange them to make them equal to each other:
$$
\(\begin{aligned}
& A D^2=A B^2-B D^2 \\
& A D^2=A E^2-E D^2
\end{aligned}\)
$$
So $\(A B^2-B D^2=A E^2-E D^2\)$.
Since we know that $\(B D>E D\)$ (and that they are both positive values), we know that $\(B D^2> E D^2\)$. In order for the above equality to hold, it follows that $\(A B^2>A E^2\)$, and therefore that $\(A B >A E\)$. The correct answer is (B).