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

Question

Accepted Answer

# Explanation:
## Step1: Use angle bisector definition
Since $ \overline{EF} $ bisects $ \angle CEB $, $ m\angle CEF = m\angle FEB $. So we set the two expressions equal:
$ 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 CEF $ or $ m\angle FEB $
Substitute $ x = 8 $ into $ m\angle CEF = 7x + 21 $:
$ m\angle CEF = 7(8) + 21 = 56 + 21 = 77^\circ $
Since $ \angle CEF = \angle FEB $, $ m\angle FEB = 77^\circ $, so $ m\angle CEB = 77^\circ + 77^\circ = 154^\circ $

## Step4: Find $ m\angle DEB $
$ \angle CEB $ and $ \angle DEB $ are supplementary (they form a linear pair), so $ m\angle DEB = 180^\circ - m\angle CEB $
$ m\angle DEB = 180 - 154 = 26^\circ $? Wait, no, wait. Wait, actually, looking at the diagram, $ \angle CEB $ and $ \angle AED $ are vertical angles, but $ \angle DEB $ and $ \angle CEA $ are vertical? Wait, no, let's re-examine. Wait, when $ EF $ bisects $ \angle CEB $, so $ \angle CEF = \angle FEB $. Then, $ \angle CEB $ is split into two equal angles. Now, $ \angle DEB $: Wait, maybe I made a mistake. Wait, actually, $ \angle CEB $ and $ \angle AED $ are vertical angles, but $ \angle DEB $ and $ \angle CEA $ are vertical? Wait, no, let's look at the lines. Lines $ AB $ and $ CD $ intersect at $ E $, so $ \angle CEB $ and $ \angle AED $ are vertical, $ \angle CEA $ and $ \angle DEB $ are vertical. Wait, but $ EF $ bisects $ \angle CEB $, so $ \angle CEF = \angle FEB $. So first, solve for $ x $:

$ 7x + 21 = 10x - 3 $

$ 21 + 3 = 10x - 7x $

$ 24 = 3x $

$ x = 8 $

Then $ m\angle CEF = 7(8) + 21 = 56 + 21 = 77 $, so $ m\angle FEB = 77 $, so $ m\angle CEB = 77 + 77 = 154 $. Now, $ \angle CEB $ and $ \angle DEB $: Wait, no, $ \angle CEB $ and $ \angle DEB $ – wait, $ CD $ is a straight line? Wait, the diagram: $ A $ and $ B $ are on a line, $ C $ and $ D $ are on another line, intersecting at $ E $. So $ \angle CEB $ and $ \angle DEB $: Wait, no, $ \angle CED $ is a straight line? Wait, maybe $ \angle DEB $ is supplementary to $ \angle CEB $? Wait, no, if $ CD $ is a straight line, then $ \angle CEB + \angle DEB = 180^\circ $? Wait, no, $ \angle CEB $ and $ \angle AED $ are vertical, $ \angle CEA $ and $ \angle DEB $ are vertical. Wait, maybe I messed up the angle. Wait, let's re-express:

Since $ EF $ bisects $ \angle CEB $, $ \angle CEF = \angle FEB $, so $ 7x + 21 = 10x - 3 $, solving gives $ x = 8 $, so $ \angle CEF = 77^\circ $, $ \angle FEB = 77^\circ $, so $ \angle CEB = 154^\circ $. Now, $ \angle DEB $: since $ \angle CEB $ and $ \angle DEB $ – wait, $ \angle CEB $ and $ \angle DEB $ are adjacent angles forming a linear pair with $ \angle CED $? Wait, no, the lines $ AB $ and $ CD $ intersect at $ E $, so $ \angle CEB $ and $ \angle AED $ are vertical angles, $ \angle CEA $ and $ \angle DEB $ are vertical angles. Wait, maybe the diagram shows that $ \angle DEB $ is vertical to $ \angle CEA $, but $ \angle CEB $ is 154°, so $ \angle CEA = 180° - 154° = 26° $, so $ \angle DEB = 26° $? Wait, no, that can't be. Wait, maybe I made a mistake in the angle relationship. Wait, let's check again.

Wait, the problem says "find the measure of $ \angle DEB $". Let's see: $ EF $ bisects $ \angle CEB $, so $ \angle CEF = \angle FEB $. So $ 7x + 21 = 10x - 3 $, so $ 3x = 24 $, $ x = 8 $. Then $ \angle CEF = 7*8 +21 = 77 $, $ \angle FEB = 77 $, so $ \angle CEB = 154 $. Now, $ \angle DEB $: since $ \angle CEB $ and $ \angle DEB $ – wait, if $ CD $ is a straight line, then $ \angle CEB + \angle DEB = 180 $? No, $ \angle CEB $ is at point $ E $, between $ C $ and $ B $, and $ D $ is on the other side of $ E $ from $ C $. So $ \angle CED $ is a straight line, so $ \angle CEB + \angle DEB = 180 $? Wait, no, $ \angle CEB $ and $ \angle DEB $ share the side $ EB $, and $ C $ and $ D $ are on a straight line, so yes, $ \angle CEB + \angle DEB = 180^\circ $. Wait, but that would mean $ \angle DEB = 180 - 154 = 26^\circ $. But that seems small. Wait, maybe the diagram is different. Wait, maybe $ \angle DEB $ is vertical to $ \angle CEA $, and $ \angle CEA $ is supplementary to $ \angle CEB $. So $ \angle CEA = 180 - 154 = 26 $, so $ \angle DEB = 26 $. But let's confirm.

Wait, another approach: since $ 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 $, so $ \angle FEB = 77 $, so $ \angle CEB = 77 + 77 = 154 $. Now, $ \angle DEB $: since $ \angle CEB $ and $ \angle DEB $ are adjacent angles forming a linear pair (because $ CD $ is a straight line), so their sum is $ 180^\circ $. Therefore, $ \angle DEB = 180 - 154 = 26^\circ $. Wait, but that seems correct? Wait, no, maybe I mixed up the angles. Wait, maybe $ \angle DEB $ is equal to $ \angle CEA $, which is supplementary to $ \angle CEB $. So yes, $ \angle CEA = 180 - 154 = 26 $, so $ \angle DEB = 26 $. But let's check again.

Wait, maybe the diagram shows that $ AB $ and $ CD $ intersect at $ E $, so $ \angle CEB $ and $ \angle AED $ are vertical angles, $ \angle CEA $ and $ \angle DEB $ are vertical angles. So $ \angle CEB = 154^\circ $, so $ \angle AED = 154^\circ $, and $ \angle CEA = 180 - 154 = 26^\circ $, so $ \angle DEB = 26^\circ $. Yes, that makes sense. So the measure of $ \angle DEB $ is $ 26^\circ $? Wait, no, wait, that can't be. Wait, maybe I made a mistake in the angle bisector. Wait, $ EF $ bisects $ \angle CEB $, so $ \angle CEF = \angle FEB $, so $ 7x + 21 = 10x - 3 $, so $ x = 8 $, then $ \angle CEF = 77 $, $ \angle FEB = 77 $, so $ \angle CEB = 154 $. Then $ \angle DEB $: if $ \angle CEB $ and $ \angle DEB $ are supplementary, then $ 180 - 154 = 26 $. So the answer is $ 26^\circ $? Wait, but let me check again.

Wait, maybe the diagram is such that $ \angle DEB $ is actually equal to $ \angle CEB $'s supplement, but maybe I messed up the lines. Alternatively, maybe $ \angle DEB $ is vertical to $ \angle CEA $, which is supplementary to $ \angle CEB $. So yes, $ 180 - 154 = 26 $. So the measure of $ \angle DEB $ is $ 26^\circ $. Wait, but that seems small. Wait, maybe I made a mistake in the equation. Let's re-solve the equation:

$ 7x + 21 = 10x - 3 $

Subtract $ 7x $ from both sides: $ 21 = 3x - 3 $

Add 3 to both sides: $ 24 = 3x $

Divide by 3: $ x = 8 $. That's correct.

Then $ \angle CEF = 7*8 +21 = 56 +21 = 77 $, $ \angle FEB = 10*8 -3 = 80 -3 = 77 $. So $ \angle CEB = 77 +77 = 154 $. Then $ \angle DEB $: since $ \angle CEB $ and $ \angle DEB $ are adjacent and form a linear pair (because $ CD $ is a straight line), their sum is $ 180^\circ $. So $ \angle DEB = 180 - 154 = 26^\circ $. Yes, that's correct.

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

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Use angle bisector definition

Step2: Solve for \( x \)

Step3: Find \( m\angle CEF \) or \( m\angle FEB \)

Step4: Find \( m\angle DEB \)

Answer: