extra credit: complete the two column proof!
as we get further into explaining our reasoning, we will start to do two column proofs with the
left column being our claims and the right column being our reasons for our claims.
given: line / is a straight line
ce bisects angle acd and cf bisects angle dcb
a = ace and a = ecd
b = dcf and b = fcb
prove: ecf a right angle (90°)

Question

Accepted Answer

# Explanation:
## Step1: Identify bisected angles
Since $CE$ bisects $\angle ACD$, $\angle ACE\cong\angle ECD$ (given $a = ACE$ and $a=ECD$). Since $CF$ bisects $\angle DCB$, $\angle DCF\cong\angle FCB$ (given $b = DCF$ and $b = FCB$). So the first blanks are $\angle ACE$ and $\angle DCF$.
## Step2: Find $\angle ACB$
A straight - line forms a $180^{\circ}$ angle. So $\angle ACB = 180^{\circ}$.
## Step3: Express $\angle ACB$ in terms of sub - angles
$\angle ACE+\angle ECD+\angle DCF+\angle FCB = 180^{\circ}$ (Angle - addition postulate and straight - line).
## Step4: Substitute $a$ and $b$
Substituting $a$ for $\angle ACE$ and $\angle ECD$, and $b$ for $\angle DCF$ and $\angle FCB$, we get $a + a + b + b=180^{\circ}$.
## Step5: Simplify
Dividing $a + a + b + b = 180^{\circ}$ (i.e., $2a + 2b=180^{\circ}$) by 2 gives $a + b = 90^{\circ}$.
## Step6: Relate to $\angle ECD+\angle DCF$
Since $a=\angle ECD$ and $b = \angle DCF$, $\angle ECD+\angle DCF=90^{\circ}$.
## Step7: Find $\angle ECF$
$\angle ECF=\angle ECD+\angle DCF$ (angle - addition postulate, combining $\angle ECD$ and $\angle DCF$ along a shared line segment). So $\angle ECF = 90^{\circ}$.

# Answer:
| Claim | Reasons |
|---|---|
|$\angle ACE\cong\angle ECD$ and $\angle DCF\cong\angle FCB$| Bisected angles |
|$\angle ACB = 180^{\circ}$| Straight line |
|$\angle ACE+\angle ECD+\angle DCF+\angle FCB = 180^{\circ}$| Angle - addition postulate and straight line |
|$a + a + b + b=180^{\circ}$| Given on a straight line with bisected angles |
|$a + b = 90^{\circ}$| Simplifying above |
|$\angle ECD+\angle DCF=90^{\circ}$| Bisected angles of two complimentary angles on a straight line |
|$\angle ECF = 90^{\circ}$| Combining $\angle ECD$ and $\angle DCF$ along a shared line segment |

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QUESTION IMAGE

Question

Explanation:

Step1: Identify bisected angles

Step2: Find \(\angle ACB\)

Step3: Express \(\angle ACB\) in terms of sub - angles

Step4: Substitute \(a\) and \(b\)

Step5: Simplify

Step6: Relate to \(\angle ECD+\angle DCF\)

Step7: Find \(\angle ECF\)

Answer:

Claim	Reasons
\(\angle ACB = 180^{\circ}\)	Straight line
\(\angle ACE+\angle ECD+\angle DCF+\angle FCB = 180^{\circ}\)	Angle - addition postulate and straight line
\(a + a + b + b=180^{\circ}\)	Given on a straight line with bisected angles
\(a + b = 90^{\circ}\)	Simplifying above
\(\angle ECD+\angle DCF=90^{\circ}\)	Bisected angles of two complimentary angles on a straight line
\(\angle ECF = 90^{\circ}\)	Combining \(\angle ECD\) and \(\angle DCF\) along a shared line segment