Question

1. the diagram on the right is a straightedge and compass construction. c is the center of both circles.

in the list below, mark all statements that must be true by construction.
segments ab and ad have the same length.
segments ac and ad have the same length.
segments ac and cd have the same length.
triangle bce is isosceles.
triangle cde is isosceles.

1. consider the situation described below.

\line fp is a perpendicular bisector of segment mh. the line and the segment intersect at point z\
a) sketch a picture of the situation described above.
b) which of the following statements must be true for this situation? (mark one)
point z is closer to point m than it is to point h.
point z is closer to point h than it is to point m.
point z is the same distance from both points m and h.
there is not enough information to be sure.
triangle abc below is isosceles. use straightedge and compass moves to construct the perpendicular bisector of segment ab. (show all your work).

Explanation:

1.

Step1: Recall properties of circles

Points on a circle are equidistant from the center. Since \(C\) is the center of both circles, \(AC = BC=CD\). And \(AD\) is a radius - like segment of the outer - circle with center \(C\), so \(AC = AD\).

Step2: Analyze isosceles triangles

In \(\triangle BCE\), \(BC = CE\) (radii of the outer - circle), so \(\triangle BCE\) is isosceles. In \(\triangle CDE\), \(CD = CE\) (radii of the outer - circle), so \(\triangle CDE\) is isosceles. \(AB\) and \(AD\) are not necessarily equal.

The perpendicular bisector of a line segment is the set of all points that are equidistant from the endpoints of the line segment. Since \(Z\) lies on the perpendicular bisector of \(MH\), by the definition of a perpendicular bisector, \(ZM=ZH\).

Answer:

• Segments \(AC\) and \(AD\) have the same length.
• Segments \(AC\) and \(CD\) have the same length.
• Triangle \(BCE\) is isosceles.
• Triangle \(CDE\) is isosceles.

2.

a)

Draw a line segment \(MH\). Then draw a line \(FP\) that intersects \(MH\) at a right - angle at the mid - point \(Z\) of \(MH\).

b)