Question

1. as shown in the figure, in square abcd, e is on ac. if ab = ae, then m∠ebc = a. 45° b. 30° c. 22.5° d. 20° 2. as shown in the figure, square abcd has a side length of 4. if be = 5, then the length of ce is a. √5 b. 5 c. √15 d. √17

Explanation:

Step1: Recall properties of a square

In square \(ABCD\), \(\angle ABC = 90^{\circ}\), and the diagonal \(AC\) bisects \(\angle BCD\) and \(\angle BAD\), so \(\angle BAC=45^{\circ}\).

Step2: Use the isosceles - triangle property

Since \(AB = AE\), in \(\triangle ABE\), \(\angle ABE=\angle AEB\). Using the angle - sum property of a triangle (\(\angle BAE + \angle ABE+\angle AEB = 180^{\circ}\)) and \(\angle BAE = 45^{\circ}\), we have \(\angle ABE=\angle AEB=\frac{180^{\circ}-45^{\circ}}{2}=67.5^{\circ}\).

Step3: Calculate \(\angle EBC\)

\(\angle EBC=\angle ABC-\angle ABE\). Since \(\angle ABC = 90^{\circ}\) and \(\angle ABE = 67.5^{\circ}\), then \(\angle EBC=90^{\circ}-67.5^{\circ}=22.5^{\circ}\).

Step1: Apply the Pythagorean theorem in square \(ABCD\)

In square \(ABCD\) with side - length \(a = 4\), in right - triangle \(BCE\), \(BC = 4\) and \(BE = 5\). According to the Pythagorean theorem \(BE^{2}=BC^{2}+CE^{2}\).

Step2: Solve for \(CE\)

We can rewrite the Pythagorean formula as \(CE=\sqrt{BE^{2}-BC^{2}}\). Substitute \(BE = 5\) and \(BC = 4\) into the formula: \(CE=\sqrt{5^{2}-4^{2}}=\sqrt{25 - 16}=\sqrt{9}=3\). But it seems there is a mistake in the options. Let's assume we use the right - triangle formed in another way. In right - triangle \(ABE\), \(AB = 4\), \(BE = 5\), then \(AE=\sqrt{BE^{2}-AB^{2}}=\sqrt{25 - 16}=3\). Since \(AD = 4\), \(DE=4 - 3 = 1\). In right - triangle \(DCE\), \(DC = 4\), \(DE = 1\), then \(CE=\sqrt{DC^{2}+DE^{2}}=\sqrt{4^{2}+1^{2}}=\sqrt{16 + 1}=\sqrt{17}\).

Answer:

C. \(22.5^{\circ}\)