Sorry, I’m new here. And so is this question to me. It would have been helpful. No, it’s not necessary if you have it memorized. Yes I get what you’re saying they are the same. I did not know that, but I do now. Thank you for taking the time to post the question though

Posted from my mobile device

no worries

These are ALWAYS the same will never change

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

You should have in mind by heart. your focus is in the question.

The above should come naturally to you to answer

can someone please explain the answer

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).

Where is all this information given all we can see is a figure of triangle and two quantities nothing else