Question

de = ce
ac = ab
df = ac

Explanation:

Step1: Analyze the figure (rectangle properties)

The figure is a rectangle \(ABCD\) (since \(AB\) and \(CD\) are opposite sides, \(AD\) and \(BC\) are opposite sides, and all angles are right angles). In a rectangle, the diagonals are equal and bisect each other. So diagonals \(AC\) and \(BD\) intersect at \(E\), which means \(E\) is the midpoint of both \(AC\) and \(BD\). So \(AE = EC\) and \(BE = ED\), and also \(AC = BD\). Now let's check each option:

• Option 1: \(DE = CE\)

In a rectangle, diagonals bisect each other, so \(DE=\frac{BD}{2}\) and \(CE = \frac{AC}{2}\). But since \(AC = BD\) (diagonals of rectangle are equal), then \(\frac{BD}{2}=\frac{AC}{2}\), so \(DE = CE\)? Wait, no, wait. Wait, diagonals \(AC\) and \(BD\) intersect at \(E\), so \(AE = EC=\frac{AC}{2}\) and \(BE = ED=\frac{BD}{2}\). Since \(AC = BD\), then \(AE = EC = BE = ED\)? Wait, no, in a rectangle, diagonals are equal and bisect each other, so \(AE = EC\) and \(BE = ED\), and \(AC = BD\), so \(AE = EC = BE = ED\). Wait, but let's check the other options.

• Option 2: \(AC = AB\)

\(AC\) is a diagonal, \(AB\) is a side. In a rectangle, diagonal is longer than the side (by Pythagoras: \(AC=\sqrt{AB^2 + BC^2}\), so \(AC>AB\) since \(BC>0\)), so \(AC
eq AB\).

• Option 3: (Assuming the third option is \(DF = AC\), but maybe typo, but let's focus on first two. Wait, the first option: \(DE = CE\). Since diagonals bisect each other and are equal, \(DE=\frac{BD}{2}\), \(CE=\frac{AC}{2}\), and \(AC = BD\), so \(DE = CE\) is true? Wait, no, wait, \(DE\) is part of diagonal \(BD\), \(CE\) is part of diagonal \(AC\). Since \(AC = BD\), then \(\frac{BD}{2}=\frac{AC}{2}\), so \(DE = CE\). Wait, but let's re - check. Wait, in a rectangle, diagonals are equal and bisect each other, so \(AE = EC\) and \(BE = ED\), and \(AC = BD\), so \(AE = EC = BE = ED\). So \(DE = CE\) because \(DE=\frac{BD}{2}\), \(CE=\frac{AC}{2}\), and \(AC = BD\), so \(DE = CE\).

Wait, but let's check the other options:

• \(AC = AB\): As \(AC\) is diagonal, \(AB\) is side, \(AC=\sqrt{AB^2 + BC^2}\), so \(AC>AB\), so \(AC

eq AB\).

So the correct option should be the first one: \(DE = CE\) (assuming the options are as given: first option \(DE = CE\), second \(AC = AB\), third maybe typo but first option is correct based on rectangle diagonal properties. Wait, maybe I made a mistake. Wait, in a rectangle, diagonals bisect each other, so \(E\) is midpoint of \(AC\) and \(BD\), so \(AE = EC\) and \(BE = ED\), and \(AC = BD\), so \(AE = EC = BE = ED\). Therefore, \(DE = CE\) (since \(DE = BE = AE = CE\)? Wait, no, \(DE\) is from \(D\) to \(E\), \(CE\) is from \(C\) to \(E\). Since \(E\) is midpoint of \(BD\) and \(AC\), and \(AC = BD\), then \(DE=\frac{BD}{2}\), \(CE=\frac{AC}{2}\), and since \(AC = BD\), then \(DE = CE\). So the first option \(DE = CE\) is correct.

Step2: Eliminate other options

• \(AC = AB\): As \(AC\) is diagonal, \(AB\) is side, \(AC=\sqrt{AB^2 + BC^2}\), so \(AC>AB\), so this is false.

• The third option (typo, but assuming) is not relevant as first option is correct.

Answer:

The correct option is the first one: \(DE = CE\) (the option with \(DE = CE\) among the given choices).