Question

2.5 notetaking with vocabulary (continued)

1. given ∠aeb is a complement of ∠bec. prove m∠aed = 90°

statements

1. ∠aeb is a complement of ∠bec.

2.

1. m∠aec = m∠aeb + m∠bec
2. m∠aec = 90°
3. m∠aed + m∠aec = 180°

6.

1. m∠aed = 90°

reasons

1. given
2. definition of complementary angles

3.

1. definition of supplementary angles
2. definition of equality
3. substitution property of equality

7.
in exercises 3 and 4, name the property that the statement illustrates.

1. if ∠rst ≅ ∠tsu and ∠tsu ≅ ∠vwx, then ∠rst ≅ ∠vwx.
2. if (overline{gh}congoverline{jk}), then (overline{jk}congoverline{gh}).
3. write a two - column proof.

given m is the midpoint of (overline{rt}). prove mt = rs + sm
statements
reasons

Explanation:

Step1: Recall definition of complementary angles

If $\angle AEB$ is a complement of $\angle BEC$, then $m\angle AEB + m\angle BEC=90^{\circ}$

Step2: Recall angle - addition postulate

The measure of $\angle AEC$ is the sum of the measures of $\angle AEB$ and $\angle BEC$, so $m\angle AEC=m\angle AEB + m\angle BEC$

Step3: Substitute from Step1 into Step2

Since $m\angle AEB + m\angle BEC = 90^{\circ}$, then $m\angle AEC = 90^{\circ}$

Step4: Recall definition of supplementary angles

$\angle AED$ and $\angle AEC$ are supplementary, so $m\angle AED+m\angle AEC = 180^{\circ}$

Step5: Substitute $m\angle AEC = 90^{\circ}$ into the supplementary - angle equation

$m\angle AED+90^{\circ}=180^{\circ}$

Step6: Solve for $m\angle AED$ using subtraction property of equality

$m\angle AED=180^{\circ}-90^{\circ}=90^{\circ}$

If $\angle RST\cong\angle TSU$ and $\angle TSU\cong\angle VWX$, then by the transitive property of congruence (if $a = b$ and $b = c$, then $a = c$ for angles in terms of congruence), $\angle RST\cong\angle VWX$

If $\overline{GH}\cong\overline{JK}$, then $\overline{JK}\cong\overline{GH}$ by the symmetric property of congruence (if $a = b$, then $b = a$ for line - segments in terms of congruence)

Step1: State the definition of a mid - point

Since $M$ is the mid - point of $\overline{RT}$, then $RM = MT$ (by the definition of a mid - point, which divides a line - segment into two equal parts)

Step2: Use segment addition postulate

We know that $RM=RS + SM$ (by the segment addition postulate, which states that if $S$ is between $R$ and $M$, then $RM=RS + SM$)

Step3: Substitute

Since $RM = MT$ and $RM=RS + SM$, then $MT=RS + SM$ (by the substitution property of equality)

Answer:

$m\angle AED = 90^{\circ}$

For exercise 3: