In the figure given below, line l is perpendicular to line m.

(a) Is CE = EG?

(b) Does $\overleftrightarrow{PE}\space$bisect segment $\overline{BH}\space$?

(c) Identify any two line segments for which $\overleftrightarrow{PE}\space$is the perpendicular bisector.

(d) Are these true?

(i) AC > FG

(ii) CD = GH

(iii) BC < EG

From the figure, the points are at positions: A(1), B(2), C(3), D(4), E(5), F(6), G(7), H(8) on line l, and line m is perpendicular to line l at point E.

(a) CE = 5 - 3 = 2 units

EG = 7 - 5 = 2 units

Since CE = EG = 2 units,

∴ Yes, CE = EG.

(b) BE = 5 - 2 = 3 units

EH = 8 - 5 = 3 units

Since BE = EH, point E is the mid-point of $\overline{BH}$. Also, $\overleftrightarrow{PE}\space$is perpendicular to line l.

∴ Yes, $\overleftrightarrow{PE}\space$bisects segment $\overline{BH}$.

(c) $\overleftrightarrow{PE}\space$is perpendicular bisector of any line segment whose midpoint is E and which lies along line l.

For example:

(i) $\overline{DF}$, since DE = EF = 1 unit and $\overleftrightarrow{PE} \perp \overline{DF}$.

(ii) $\overline{BH}$, since BE = EH = 3 units and $\overleftrightarrow{PE} \perp \overline{BH}$.

∴ $\overleftrightarrow{PE}\space$is the perpendicular bisector of $\overline{DF}$ and $\overline{BH}$.

(d) (i) AC = 3 - 1 = 2 units

FG = 7 - 6 = 1 unit

Since 2 > 1,

∴ AC > FG is True.

(ii) CD = 4 - 3 = 1 unit

GH = 8 - 7 = 1 unit

Since CD = GH,

∴ CD = GH is True.

(iii) BC = 3 - 2 = 1 unit

EG = 7 - 5 = 2 units

Since 1 < 2,

∴ BC < EG is True.