for numbers 10 and 11, find each measure.
10.
(there is a diagram of a quadrilateral abcd with markings: ab and ad are marked equal, bc and dc are marked equal, angle at a is 31 degrees, and there is a dashed line from b to d and from a to c intersecting at the center. also, there is some handwritten text near a and some scribbles at the bottom right.)

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Accepted Answer

# Explanation:
## Step1: Identify the figure type
The figure is a kite (or a rhombus - like figure with two pairs of adjacent sides equal, as seen from the tick marks: $AB = AD$ and $BC = DC$, and the diagonals intersecting). In a kite, one diagonal bisects the vertex angles, and the diagonals are perpendicular? Wait, actually, in a kite with two pairs of adjacent sides equal, the diagonal $AC$ bisects $\angle A$ and $\angle C$, and diagonal $BD$ is the axis of symmetry. Also, triangle $ABC$ and $ADC$ are congruent? Wait, no, $AB = AD$, $BC = DC$, $AC$ is common, so $\triangle ABC\cong\triangle ADC$ by SSS. But also, $AB = AD$, so $\triangle ABD$ is isosceles with $AB = AD$. Wait, the angle at $A$ is $31^\circ$, and the diagonal $BD$ is split by $AC$ into two right angles? Wait, no, actually, in a kite, one diagonal is perpendicular to the other? Wait, maybe it's a rhombus? No, the tick marks: $AB$ and $AD$ have one tick, $BC$ and $DC$ have two ticks, so $AB = AD$, $BC = DC$, so it's a kite with $AB = AD$, $BC = DC$. Then, diagonal $AC$ bisects $\angle A$ and $\angle C$, and diagonal $BD$ is bisected by $AC$ at right angles? Wait, maybe we can consider triangle $ABO$ (where $O$ is the intersection of diagonals) as a right triangle? Wait, maybe the key is that in triangle $ABD$, $AB = AD$, so it's isosceles, and $AC$ is the angle bisector, so it's also the altitude and median. Wait, the angle at $A$ is $31^\circ$, so in triangle $ABD$, which is isosceles with $AB = AD$, the base angles at $B$ and $D$ would be equal. But we need to find $\angle 1$ (angle at $B$ between $AB$ and $BD$). Wait, maybe the diagonals are perpendicular? Let's assume that the diagonals $AC$ and $BD$ are perpendicular (a property of a kite: one diagonal is perpendicular to the other). So, in triangle $ABO$ (where $O$ is the intersection of $AC$ and $BD$), $\angle AOB = 90^\circ$, $\angle OAB = \frac{31^\circ}{2}$? Wait, no, the angle at $A$ is $31^\circ$, and $AC$ bisects it? Wait, the diagram shows angle at $A$ is $31^\circ$, and $AC$ is a dashed line, maybe bisecting the angle? Wait, no, the marks on $AB$ and $AD$ are the same, so $AB = AD$, so triangle $ABD$ is isosceles with $AB = AD$, so $\angle ABD=\angle ADB$. The sum of angles in a triangle is $180^\circ$, so in triangle $ABD$, $\angle BAD = 31^\circ$, so $\angle ABD+\angle ADB = 180 - 31=149^\circ$, and since $\angle ABD=\angle ADB$, each is $74.5^\circ$? Wait, but maybe the diagonals are perpendicular, so $\angle 1$ is $90^\circ - \frac{31^\circ}{2}$? No, wait, maybe I made a mistake. Wait, the figure: $AB = AD$ (one tick), $BC = DC$ (two ticks), $AC$ is common, so $\triangle ABC\cong\triangle ADC$ (SSS). Then, $AC$ bisects $\angle BCD$ and $\angle BAD$. So $\angle BAC=\angle DAC = \frac{31^\circ}{2}$? No, the angle at $A$ is $31^\circ$, so maybe $AC$ is the angle bisector, so $\angle BAC = 31^\circ$ (wait, the diagram shows angle at $A$ is $31^\circ$, between $AB$ and $AC$? Wait, looking at the diagram: point $A$, with a dashed line to $C$, and another dashed line to $B$ and $D$. The angle at $A$ between $AB$ and $AD$ is $31^\circ$? No, the angle marked $31^\circ$ is between $AB$ and $AC$? Wait, the diagram has angle at $A$ labeled $31^\circ$, between the two sides $AB$ and $AD$? No, the marks: $AB$ and $AD$ have one tick, so $AB = AD$, so triangle $ABD$ is isosceles with $AB = AD$, and $AC$ is a line from $A$ to $C$, with $BC = DC$ (two ticks). So $AC$ is the axis of symmetry? Wait, maybe the diagonals are perpendicular, so $\angle 1$ is $90^\circ - 31^\circ=59^\circ$? Wait, no, let's think again. If we consider triangle $ABO$ (where $O$ is the intersection of diagonals), if diagonals are perpendicular, then $\angle AOB = 90^\circ$, angle at $A$ is $31^\circ$, so angle at $B$ (angle $1$) is $90^\circ - 31^\circ = 59^\circ$? Wait, no, maybe the angle at $A$ is $31^\circ$, and in triangle $ABC$, which is a right triangle? Wait, maybe the key is that in a kite, one diagonal is perpendicular to the other, so the triangle formed (like $\triangle AOB$) is a right triangle. So if angle at $A$ is $31^\circ$, then angle $1$ (at $B$) is $90^\circ - 31^\circ = 59^\circ$? Wait, no, let's check the sum of angles in a triangle. In triangle $ABO$, if $\angle AOB = 90^\circ$, $\angle OAB = 31^\circ$, then $\angle OBA=\angle 1 = 90 - 31=59^\circ$. Wait, but maybe the angle at $A$ is $31^\circ$, and $AB = AD$, so triangle $ABD$ is isosceles, so the base angles are equal. Wait, maybe I misread the angle. Let's re - examine: the diagram shows angle at $A$ is $31^\circ$, between $AB$ and $AC$, and $AC$ is perpendicular to $BD$. So in right triangle $ABC$ (wait, $AC\perp BD$), so $\angle 1$ is $90^\circ - 31^\circ = 59^\circ$. Wait, but let's confirm: in a kite, one diagonal is perpendicular to the other, so $\angle AOB = 90^\circ$ (where $O$ is the intersection of $AC$ and $BD$). Then, in triangle $AOB$, $\angle OAB = 31^\circ$, $\angle AOB = 90^\circ$, so $\angle OBA=\angle 1 = 180 - 90 - 31=59^\circ$.

## Step2: Calculate angle 1
Using the right - triangle angle sum property ($180^\circ$ in a triangle), in $\triangle AOB$ (assuming $\angle AOB = 90^\circ$ because diagonals in a kite are perpendicular), we have:
$\angle 1=180^\circ - 90^\circ - 31^\circ = 59^\circ$

# Answer:
$59^\circ$

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