29
type the correct answer in each box.
quadrilateral ebcd is an isosceles trapezoid with ( mangle edc = 110^{circ}, mangle abc = 133^{circ} ), and ( mangle dea = 114^{circ} ).
( mangle deb=square )
( mangle bcd=square )
( mangle eab=square )

Question

Accepted Answer

# Explanation:
## Step1: Find $ m\angle DEB $
In an isosceles trapezoid $ EBCD $, $ DE \parallel BC $? Wait, no, in an isosceles trapezoid, the base angles are equal and consecutive angles between the bases are supplementary. Wait, $ EBCD $ is isosceles trapezoid with $ ED $ and $ BC $ as the legs? Wait, no, the sides with ticks are $ ED $ and $ BC $, so $ ED = BC $, and $ EB \parallel DC $ (since it's a trapezoid, one pair of sides is parallel). So $ EB \parallel DC $, so consecutive angles between them are supplementary. So $ \angle EDC + \angle DEB = 180^\circ $? Wait, no, $ \angle EDC $ and $ \angle DEB $: if $ EB \parallel DC $, then $ \angle EDC + \angle DEB = 180^\circ $? Wait, $ \angle EDC = 110^\circ $, so $ \angle DEB = 180^\circ - 110^\circ = 70^\circ $? Wait, no, maybe I got the parallel sides wrong. Wait, in trapezoid $ EBCD $, the two parallel sides are $ EB $ and $ DC $, so $ EB \parallel DC $. Then $ \angle EDC $ and $ \angle DEB $: are they same - side interior angles? Let's see, $ ED $ is a transversal. So $ \angle EDC + \angle DEB = 180^\circ $, so $ \angle DEB = 180 - 110 = 70^\circ $. Wait, but let's check another way. Wait, maybe $ \angle DEB $ is adjacent to $ \angle DEA $. Wait, $ \angle DEA = 114^\circ $, and $ \angle DEA + \angle DEB = 180^\circ $? No, that would be if $ AEB $ is a straight line, which it is, because $ E $, $ B $ are connected, and $ A $ is a vertex. Wait, $ \angle DEA = 114^\circ $, so $ \angle DEB = 180^\circ - 114^\circ = 66^\circ $? Wait, no, that's not right. Wait, maybe I messed up the diagram. Let's re - examine. The quadrilateral $ EBCD $ is isosceles trapezoid, so $ ED = BC $, and $ EB \parallel DC $. So $ \angle EDC + \angle DEB = 180^\circ $ (same - side interior angles). So $ \angle DEB = 180 - 110 = 70^\circ $. Wait, but let's check the other angle. Wait, maybe the first step is to find $ \angle DEB $. Let's confirm: in isosceles trapezoid, base angles are equal. So $ \angle EDC = \angle BCD $? Wait, no, $ \angle EDC $ and $ \angle BCD $: if $ EB \parallel DC $, then $ \angle EDC + \angle DEB = 180^\circ $, and $ \angle BCD + \angle EBC = 180^\circ $, and since it's isosceles, $ \angle DEB=\angle EBC $, $ \angle EDC=\angle BCD $. Wait, $ \angle EDC = 110^\circ $, so $ \angle BCD = 110^\circ $? No, that can't be, because in a trapezoid, consecutive angles between the bases are supplementary. Wait, I think I made a mistake. Let's start over.

1. For $ m\angle DEB $:
Since $ EBCD $ is an isosceles trapezoid with $ EB\parallel DC $, the same - side interior angles $ \angle EDC $ and $ \angle DEB $ are supplementary.
So $ m\angle DEB=180^{\circ}-m\angle EDC $
Given $ m\angle EDC = 110^{\circ} $, then $ m\angle DEB = 180 - 110=70^{\circ} $

2. For $ m\angle BCD $:
In an isosceles trapezoid, base angles are equal. Also, since $ EB\parallel DC $, $ \angle EDC $ and $ \angle BCD $: Wait, no, $ \angle EDC $ and $ \angle BCD $ are not same - side interior angles. Wait, $ \angle EDC $ and $ \angle BCD $: in isosceles trapezoid $ EBCD $ with $ ED = BC $ and $ EB\parallel DC $, $ \angle EDC=\angle BCD $? No, that's not correct. Wait, consecutive angles between the legs and the base: $ \angle EDC + \angle BCD=180^{\circ} $? No, that's not right. Wait, in a trapezoid, if $ AB \parallel CD $, then $ \angle A+\angle D = 180^{\circ} $, $ \angle B+\angle C = 180^{\circ} $. So in trapezoid $ EBCD $ with $ EB\parallel DC $, $ \angle EDC+\angle DEB = 180^{\circ} $ and $ \angle BCD+\angle EBC = 180^{\circ} $. And since it's isosceles, $ \angle DEB=\angle EBC $, so $ \angle EDC=\angle BCD $. Wait, that makes sense. So $ m\angle BCD=m\angle EDC = 110^{\circ} $

3. For $ m\angle EAB $:
First, in triangle $ AEB $, we know two angles? Wait, we know $ m\angle DEA = 114^{\circ} $, $ m\angle DEB = 70^{\circ} $, so $ m\angle AEB=m\angle DEA - m\angle DEB=114 - 70 = 44^{\circ} $? No, wait, $ \angle DEA $ and $ \angle DEB $: if $ E $, $ B $ are on a straight line from $ A $'s perspective? Wait, $ \angle DEA = 114^{\circ} $, $ \angle DEB = 70^{\circ} $, so $ \angle AEB=\angle DEA-\angle DEB = 114 - 70=44^{\circ} $? Wait, no, maybe $ \angle AEB = 180^{\circ}-\angle DEA-\angle DEB $? No, that's not. Wait, $ \angle DEA = 114^{\circ} $, $ \angle DEB = 70^{\circ} $, so $ \angle AEB=180-(114 + 70)= - 4 $, which is impossible. So I must have messed up $ \angle DEB $.

Wait, let's re - consider $ \angle DEB $. Since $ EBCD $ is an isosceles trapezoid, $ ED = BC $ and $ EB\parallel DC $. So $ \angle EDC+\angle DEB = 180^{\circ} $, so $ \angle DEB = 70^{\circ} $. Then, $ \angle AEB $: $ \angle DEA = 114^{\circ} $, so $ \angle AEB=180^{\circ}-\angle DEA=180 - 114 = 66^{\circ} $? Wait, no, $ \angle DEA $ and $ \angle AEB $ are adjacent angles forming a linear pair? If $ E $, $ B $ are connected, and $ A $ is a vertex, then $ \angle DEA+\angle AEB = 180^{\circ} $, so $ \angle AEB = 180 - 114 = 66^{\circ} $. Then, $ \angle ABE $: $ \angle ABC = 133^{\circ} $, and $ \angle ABE=\angle ABC-\angle EBC $. In isosceles trapezoid $ EBCD $, $ \angle EBC=\angle DEB = 70^{\circ} $ (because in isosceles trapezoid, base angles are equal, and $ \angle DEB $ and $ \angle EBC $ are base angles). So $ \angle ABE = 133 - 70 = 63^{\circ} $. Then in triangle $ AEB $, the sum of angles is $ 180^{\circ} $. So $ m\angle EAB=180-(m\angle AEB + m\angle ABE) $
We have $ m\angle AEB = 66^{\circ} $, $ m\angle ABE = 63^{\circ} $
So $ m\angle EAB=180-(66 + 63)=180 - 129 = 51^{\circ} $

Wait, let's correct the steps:

## Step1: Find $ m\angle DEB $
Since $ EBCD $ is an isosceles trapezoid with $ EB\parallel DC $, $ \angle EDC $ and $ \angle DEB $ are same - side interior angles, so they are supplementary.
$ m\angle DEB=180^{\circ}-m\angle EDC $
Given $ m\angle EDC = 110^{\circ} $, so $ m\angle DEB = 180 - 110 = 70^{\circ} $

## Step2: Find $ m\angle BCD $
In an isosceles trapezoid, base angles are equal. Also, since $ EB\parallel DC $, $ \angle EDC $ and $ \angle BCD $: Wait, no, $ \angle EDC $ and $ \angle BCD $ are same - side interior angles? No, $ \angle EDC $ and $ \angle BCD $ are adjacent angles? Wait, no, in trapezoid $ EBCD $, $ DC $ is the base, $ ED $ and $ BC $ are the legs. So $ \angle EDC $ and $ \angle BCD $ are base angles? No, in a trapezoid, the base angles are the angles adjacent to each base. So if $ DC $ is a base, then $ \angle EDC $ and $ \angle BCD $ are base angles. In an isosceles trapezoid, base angles are equal? No, that's not correct. Wait, in an isosceles trapezoid, the base angles are equal. So if $ EB $ and $ DC $ are the two bases, then the base angles are $ \angle EDC=\angle BCD $ and $ \angle DEB=\angle EBC $. Wait, no, that's the opposite. Wait, let's use the property that in an isosceles trapezoid, consecutive angles between the legs and the base are supplementary. So $ \angle EDC+\angle DEB = 180^{\circ} $ and $ \angle BCD+\angle EBC = 180^{\circ} $, and since $ ED = BC $, $ \angle DEB=\angle EBC $, so $ \angle EDC=\angle BCD $. Therefore, $ m\angle BCD=m\angle EDC = 110^{\circ} $

## Step3: Find $ m\angle EAB $
First, find $ \angle AEB $: $ \angle DEA = 114^{\circ} $, and $ \angle DEA+\angle AEB = 180^{\circ} $ (linear pair), so $ m\angle AEB=180 - 114 = 66^{\circ} $
Then, find $ \angle ABE $: $ \angle ABC = 133^{\circ} $, and $ \angle ABE=\angle ABC-\angle EBC $. Since $ \angle EBC=\angle DEB = 70^{\circ} $ (from isosceles trapezoid property, $ \angle DEB=\angle EBC $), so $ m\angle ABE = 133 - 70 = 63^{\circ} $
In triangle $ AEB $, the sum of interior angles is $ 180^{\circ} $. So $ m\angle EAB=180-(m\angle AEB + m\angle ABE) $
Substitute $ m\angle AEB = 66^{\circ} $ and $ m\angle ABE = 63^{\circ} $
$ m\angle EAB=180-(66 + 63)=180 - 129 = 51^{\circ} $

# Answer:
$ m\angle DEB=\boxed{70} $, $ m\angle BCD=\boxed{110} $, $ m\angle EAB=\boxed{51} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Find \( m\angle DEB \)

Answer: