20. if $overrightarrow{ef}$ bisects $angle ceb$, $mangle cef = 7x + 21$ and $mangle feb = 10x - 3$, find the measure of $angle deb$.

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Accepted Answer

# Explanation:
## Step1: Use angle bisector definition
Since $ \overrightarrow{EF} $ bisects $ \angle CEB $, $ m\angle CEF = m\angle FEB $.
So, $ 7x + 21 = 10x - 3 $.

## Step2: Solve for $ x $
Subtract $ 7x $ from both sides: $ 21 = 3x - 3 $.
Add 3 to both sides: $ 24 = 3x $.
Divide by 3: $ x = 8 $.

## Step3: Find $ m\angle FEB $
Substitute $ x = 8 $ into $ 10x - 3 $: $ 10(8) - 3 = 80 - 3 = 77 $.
Since $ \angle CEF = \angle FEB = 77^\circ $, $ m\angle CEB = 77 + 77 = 154^\circ $.

## Step4: Find $ m\angle DEB $
$ \angle CEB $ and $ \angle DEB $ are supplementary (linear pair), so $ m\angle DEB = 180 - 154 = 26^\circ $? Wait, no—wait, maybe $ \angle AEB $ is straight? Wait, no, let's check again. Wait, actually, $ \angle CEB $ is bisected into two angles each $ 77^\circ $, so $ \angle CEB = 154^\circ $. Then $ \angle DEB $: Wait, maybe $ \angle AED $ and $ \angle CEB $ are vertical angles? No, looking at the diagram, $ AEB $ is a straight line? Wait, no, the diagram shows lines intersecting at E. Wait, maybe I made a mistake. Wait, let's re-express:

Wait, when $ EF $ bisects $ \angle CEB $, $ \angle CEF = \angle FEB $. So $ 7x +21 = 10x -3 $, solve for x:

$ 7x +21 = 10x -3 $

$ 21 +3 = 10x -7x $

$ 24 = 3x $

$ x = 8 $

Then $ \angle CEF = 7(8)+21 = 56 +21 = 77 $, $ \angle FEB = 10(8)-3 = 80 -3 = 77 $. So $ \angle CEB = 77 +77 = 154^\circ $. Now, $ \angle DEB $: Wait, maybe $ \angle AEC $ and $ \angle DEB $ are vertical angles? No, looking at the diagram, lines $ AB $ and $ CD $ intersect at E, so $ \angle AEC = \angle DEB $, and $ \angle CEB + \angle AEC = 180^\circ $ (linear pair). Wait, no—if $ AB $ is a straight line, then $ \angle AEB = 180^\circ $, so $ \angle CEB + \angle AEC = 180^\circ $, but $ \angle AEC = \angle DEB $ (vertical angles). Wait, no, maybe $ \angle DEB $ is supplementary to $ \angle CEB $? Wait, no, let's see: if $ CD $ and $ AB $ intersect at E, then $ \angle CEB $ and $ \angle AED $ are vertical angles, and $ \angle DEB $ and $ \angle AEC $ are vertical angles. Wait, maybe I misread the diagram. Wait, the problem is to find $ m\angle DEB $. Wait, maybe $ \angle DEB $ is equal to $ \angle AEC $, and $ \angle AEC + \angle CEB = 180^\circ $ (since $ AB $ is a straight line). So $ \angle AEC = 180 - 154 = 26^\circ $, so $ \angle DEB = 26^\circ $? Wait, but that seems small. Wait, no, maybe I made a mistake in the diagram interpretation. Wait, let's check again.

Wait, the problem says "find the measure of $ \angle DEB $". Let's re-express:

Since $ EF $ bisects $ \angle CEB $, $ \angle CEF = \angle FEB $, so $ 7x +21 = 10x -3 $, so $ x = 8 $, so $ \angle CEF = 77 $, $ \angle FEB = 77 $, so $ \angle CEB = 154 $. Now, if $ AB $ and $ CD $ are straight lines intersecting at E, then $ \angle CEB $ and $ \angle AED $ are vertical angles (so $ \angle AED = 154 $), and $ \angle DEB $ and $ \angle AEC $ are vertical angles. Also, $ \angle AEC + \angle CEB = 180^\circ $ (linear pair), so $ \angle AEC = 180 - 154 = 26^\circ $, so $ \angle DEB = 26^\circ $. Wait, but that seems correct? Wait, no, maybe the diagram is different. Wait, maybe $ \angle DEB $ is adjacent to $ \angle CEB $ as a linear pair? Wait, no, $ \angle CEB $ and $ \angle DEB $: if $ D $ is on the other side, maybe $ \angle DEB $ is supplementary to $ \angle CEB $? No, that would be $ 180 - 154 = 26 $, which is what we have. Wait, maybe I made a mistake in the angle bisector. Wait, no, angle bisector divides the angle into two equal parts, so $ \angle CEF = \angle FEB $, so solving gives $ x = 8 $, then each angle is 77, so $ \angle CEB = 154 $, then $ \angle DEB = 180 - 154 = 26 $. Wait, but let's check again.

Wait, maybe the diagram has $ AB $ and $ CD $ intersecting at E, so $ \angle AEC = \angle DEB $ (vertical angles), and $ \angle CEB + \angle AEC = 180^\circ $ (since $ AB $ is a straight line). So yes, $ \angle AEC = 180 - 154 = 26 $, so $ \angle DEB = 26^\circ $. Wait, but that seems correct. Alternatively, maybe I messed up the angle bisector. Wait, no, the angle bisector means the two angles are equal, so that step is correct. Then solving for x is correct. Then $ \angle CEB = 154 $, then $ \angle DEB = 26 $. Wait, but let's check with x=8: $ 7x+21=77 $, $ 10x-3=77 $, so that's correct. Then $ \angle CEB = 77 +77=154 $. Then $ \angle DEB $: since $ AB $ is a straight line, $ \angle AEB = 180^\circ $, so $ \angle AEC + \angle CEB = 180^\circ $, so $ \angle AEC = 26^\circ $, and $ \angle AEC = \angle DEB $ (vertical angles), so $ \angle DEB = 26^\circ $. Yes, that makes sense.

# Answer:
$ \boxed{26^\circ} $

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QUESTION IMAGE

Question

Explanation:

Step1: Use angle bisector definition

Step2: Solve for \( x \)

Step3: Find \( m\angle FEB \)

Step4: Find \( m\angle DEB \)

Answer: