question
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m∠c=

m∠d=

m∠a=

Question

question
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m∠c=

m∠d=

m∠a=

Accepted Answer

# Explanation:
## Step1: Identify the trapezoid type
The figure is an isosceles trapezoid (legs $AB$ and $CD$ are equal, and bases $BC$ and $AD$ are parallel). In an isosceles trapezoid, consecutive angles between bases are supplementary, and base angles are equal. Also, adjacent angles on the same leg are supplementary.

## Step2: Find $m\angle C$
Since $BC \parallel AD$ and $AB$ is a leg, $\angle B$ and $\angle C$ are adjacent angles on the same leg $BC$? Wait, no, actually, in a trapezoid with bases $BC$ and $AD$, the legs are $AB$ and $CD$. So $\angle B$ and $\angle A$ are supplementary (same - side interior angles), $\angle B$ and $\angle C$ are... Wait, no, let's correct. In a trapezoid, consecutive angles between the bases are supplementary. Wait, $BC \parallel AD$, so the transversal $AB$ creates same - side interior angles $\angle B$ and $\angle A$, so $\angle B+\angle A = 180^{\circ}$. The transversal $CD$ creates same - side interior angles $\angle C$ and $\angle D$, so $\angle C+\angle D=180^{\circ}$. Also, in an isosceles trapezoid, $\angle B=\angle C$? No, wait, no. Wait, in an isosceles trapezoid, base angles are equal. The bases are $BC$ and $AD$. So angles adjacent to each base are equal. So $\angle B=\angle C$? No, wait, let's look at the sides. $AB = CD$ (marked with ticks), $BC\parallel AD$. So $\angle B$ and $\angle C$: since $AB = CD$ and $BC\parallel AD$, $\angle B$ and $\angle C$ are supplementary? Wait, no, let's use the property of isosceles trapezoid: consecutive angles between the legs and the bases. Wait, actually, in a trapezoid, if $BC\parallel AD$, then $\angle B+\angle A=180^{\circ}$ (same - side interior angles), $\angle C+\angle D = 180^{\circ}$ (same - side interior angles). And in an isosceles trapezoid, $\angle A=\angle D$ and $\angle B=\angle C$? Wait, no, that's not right. Wait, let's take a step back. The given angle is $\angle B = 115^{\circ}$. Since $BC\parallel AD$, and $AB$ is a leg, $\angle B$ and $\angle A$ are same - side interior angles, so $\angle B+\angle A=180^{\circ}$, so $\angle A = 180 - 115=65^{\circ}$. Now, for $\angle C$: since $AB = CD$ (isosceles trapezoid), $\angle B$ and $\angle C$ are supplementary? Wait, no, $\angle B$ and $\angle C$: let's consider the sides $BC$ and $CD$. Wait, maybe I made a mistake. Let's recall the properties of isosceles trapezoid:

1. Base angles are equal. The two angles adjacent to each base are equal. So if the bases are $AD$ and $BC$, then $\angle A=\angle D$ and $\angle B=\angle C$ is wrong. Wait, no, actually, the base angles are the angles adjacent to each base. So for base $AD$, the base angles are $\angle A$ and $\angle D$, so $\angle A=\angle D$. For base $BC$, the base angles are $\angle B$ and $\angle C$, so $\angle B=\angle C$ is wrong. Wait, no, let's check with a standard isosceles trapezoid. In an isosceles trapezoid $ABCD$ with $AD\parallel BC$, $AB = CD$, then $\angle A+\angle B = 180^{\circ}$, $\angle C+\angle D=180^{\circ}$, and $\angle A=\angle D$, $\angle B=\angle C$. Wait, yes! Because $AD\parallel BC$, so $\angle A$ and $\angle B$ are same - side interior angles (supplementary), $\angle C$ and $\angle D$ are same - side interior angles (supplementary). And since $AB = CD$, by the isosceles trapezoid theorem, $\angle A=\angle D$ and $\angle B=\angle C$ is incorrect. Wait, no, the correct property is that in an isosceles trapezoid, each pair of base angles is equal. So if the bases are $AD$ and $BC$, then $\angle A=\angle D$ (base angles for base $AD$) and $\angle B=\angle C$ (base angles for base $BC$)? No, that can't be, because if $AD\parallel BC$, $\angle A+\angle B = 180^{\circ}$, so if $\angle B=\angle C$ and $\angle A=\angle D$, then $\angle C+\angle D=180^{\circ}$ as well.

Wait, let's start over. The figure has $BC\parallel AD$ (arrows on $BC$ and $AD$ indicate parallelism), and $AB = CD$ (ticks on $AB$ and $CD$). So it's an isosceles trapezoid.

- For $\angle C$: Since $BC\parallel AD$ and $CD$ is a leg, $\angle B$ and $\angle C$ are supplementary? No, $\angle B$ and $\angle A$ are supplementary ($BC\parallel AD$, transversal $AB$), so $\angle A=180 - 115 = 65^{\circ}$. Then, since $AB = CD$, $\angle D=\angle A = 65^{\circ}$ (base angles of isosceles trapezoid are equal). Then, $\angle C$ and $\angle D$ are supplementary ( $BC\parallel AD$, transversal $CD$), so $\angle C=180 - 65=115^{\circ}$? Wait, no, that's conflicting. Wait, no, $\angle B$ and $\angle C$: let's look at the sides. $AB$ and $CD$ are legs, $BC$ and $AD$ are bases. So $\angle B$ is adjacent to base $BC$ and leg $AB$, $\angle C$ is adjacent to base $BC$ and leg $CD$. Since $AB = CD$ and $BC\parallel AD$, $\angle B$ and $\angle C$ are supplementary? No, let's use the property of consecutive angles in a trapezoid. In a trapezoid, consecutive angles between the legs are supplementary? No, the correct property is that in a trapezoid with bases $b_1$ and $b_2$, the angles adjacent to each leg are supplementary. So for leg $AB$, the angles adjacent to it are $\angle A$ and $\angle B$, so $\angle A+\angle B = 180^{\circ}$. For leg $CD$, the angles adjacent to it are $\angle C$ and $\angle D$, so $\angle C+\angle D=180^{\circ}$. And in an isosceles trapezoid, $\angle A=\angle D$ and $\angle B=\angle C$ is wrong. Wait, no, the correct property is that in an isosceles trapezoid, each pair of base angles is equal. So base angles at base $AD$ are $\angle A$ and $\angle D$ (equal), and base angles at base $BC$ are $\angle B$ and $\angle C$ (equal). But since $BC\parallel AD$, $\angle A+\angle B = 180^{\circ}$ (same - side interior angles), so $\angle D+\angle C=180^{\circ}$ (since $\angle A=\angle D$ and $\angle B=\angle C$).

Given $\angle B = 115^{\circ}$, then:

- $\angle A=180 - 115 = 65^{\circ}$ (supplementary to $\angle B$ because $BC\parallel AD$, transversal $AB$)
- Since $AB = CD$, $\angle D=\angle A = 65^{\circ}$ (base angles of isosceles trapezoid are equal)
- $\angle C$ and $\angle D$ are supplementary ( $BC\parallel AD$, transversal $CD$), so $\angle C=180 - 65 = 115^{\circ}$
- Also, since $AB = CD$, $\angle B=\angle C = 115^{\circ}$ (base angles at base $BC$ are equal)

# Answer:
$m\angle C = 115^{\circ}$

$m\angle D=65^{\circ}$

$m\angle A = 65^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the trapezoid type

Step2: Find \(m\angle C\)

Answer: