Question

question 11 in the diagram, bh is the perpendicular bisector of pq. which statement can be proven to be true? 1. ph = qh 2. bh = ph = qh 3. /bph=/bqh 4. /pbh=/qbh=/phb

Explanation:

Step1: Recall property of perpendicular bisector

If a point lies on the perpendicular bisector of a line - segment, then it is equidistant from the endpoints of the line - segment. Here, point \(H\) lies on the perpendicular bisector \(HS\) of \(PQ\). So, \(PH = QH\).

Step2: Analyze angle - related properties

Since \(HS\) is the perpendicular bisector of \(PQ\), \(\angle PSH=\angle QSH = 90^{\circ}\), and \(PS = QS\). Triangles \(\triangle PSH\) and \(\triangle QSH\) are congruent by the Side - Angle - Side (SAS) congruence criterion (\(PS = QS\), \(\angle PSH=\angle QSH\), \(SH = SH\)). So, \(\angle SPH=\angle SQH\).

Answer:

1. \(PH\cong QH\) is true because of the property of the perpendicular bisector (a point on the perpendicular bisector of a line - segment is equidistant from the endpoints of the line - segment).
2. \(SH\cong PH\cong QH\) is not necessarily true. There is no information to suggest that \(SH\) is equal in length to \(PH\) or \(QH\).
3. \(\angle SPH=\angle SQH\) is true as \(\triangle PSH\cong\triangle QSH\) (by SAS congruence criterion: \(PS = QS\), \(\angle PSH=\angle QSH = 90^{\circ}\), \(SH\) is common).
4. \(\angle SPH=\angle SHP=\angle SHQ\) is not true. \(\angle SHP+\angle SPH = 90^{\circ}\) and \(\angle SHQ+\angle SQH = 90^{\circ}\), and in general, these angles are not equal to each other.

So, the true statements are \(PH\cong QH\) and \(\angle SPH=\angle SQH\).