Question

1. as shown in the figure, in square abcd, ae = ab. line segment de intersects with segment bc at point f. then m∠bef = ____. a. 45° b. 30° c. 60° d. 55° 2. as shown in the figure, quadrilateral abcd has its diagonals ac and bd bisecting each other. in order for it to become a square, an additional condition must be added. this condition could be ____. a. ab = cd b. ac = bd c. ac⊥bd d. ac = bd and ac⊥bd

Explanation:

Step1: Analyze triangle ABE in square ABCD

In square \(ABCD\), \(AB = AD\), and since \(AE=AB\), then \(AE = AD\). \(\angle BAD=90^{\circ}\), in \(\triangle ABE\), \(AE = AB\), so \(\angle ABE=\angle AEB\). And \(\angle BAE=\angle BAD - \angle EAD\). In \(\triangle ADE\), \(AE = AD\), so \(\angle AED=\angle ADE\). Let \(\angle EAD = x\), then \(\angle AED=\angle ADE=\frac{180^{\circ}-\angle EAD}{2}=90^{\circ}-\frac{x}{2}\), \(\angle AEB=\angle ABE=\frac{180^{\circ}-\angle BAE}{2}\), and \(\angle BAE = 90^{\circ}-x\), so \(\angle AEB = 45^{\circ}+\frac{x}{2}\).

Step2: Calculate \(\angle BEF\)

\(\angle BEF = 180^{\circ}-\angle AEB-\angle AED\). Substitute \(\angle AEB = 45^{\circ}+\frac{x}{2}\) and \(\angle AED=90^{\circ}-\frac{x}{2}\) into it. \(\angle BEF=180^{\circ}-(45^{\circ}+\frac{x}{2})-(90^{\circ}-\frac{x}{2}) = 45^{\circ}\).

For question 2: Recall square - determination conditions

A quadrilateral with diagonals bisecting each other is a parallelogram. For a parallelogram to be a square, its diagonals should be equal and perpendicular. If the diagonals \(AC\) and \(BD\) of quadrilateral \(ABCD\) bisect each other, adding the condition \(AC = BD\) and \(AC\perp BD\) can make it a square.

Answer:

1. A. \(45^{\circ}\)
2. D. \(AC = BD\) and \(AC\perp BD\)