Problem:
$AB$ and $AC$ are equal legs in $\triangle ABC$. Point $D$ lies on $AB$ such that $CD = CB$. If $\angle ADC = 114^\circ$, what is $\angle ACD$ in degrees?
BDC = 66
so, DBC = 66
BCD = 48
ACB=66, so ACD = 18
Well, isn't this triangle quite the equalizer! Are you sure it's not promoting equality? Anyway, let's get down to business.
Since $AB$ and $AC$ are equal legs, we have an isosceles triangle. Let's name the point of intersection of the legs as $O$ – or should I say, the "Center of Equality" – just in case it wants to promote its agenda.
Now, we also have $CD = CB$, which means $\triangle CBD$ is isosceles as well. Since $CB$ is equal to both $CD$ and $CB$ itself (quite redundant, isn't it?), it must be an equilateral triangle! I guess $C$ is the "Center for Balanced Distances." I must say, this triangle has quite the political message!
Moving on, we're given $\angle ADC = 114^\circ$. This angle comes out of nowhere, like a surprise math test! Considering $C$ is the center of equality, it must be very fair, so let's assume $\angle ADC$ is evenly split between $\angle ACD$ and $\angle ADB$. That sounds like something a fair and balanced $C$ would do.
So, $\angle ACD$ and $\angle ADB$ must both be $57^\circ$. Why? Because that's half of $114^\circ$, silly – just like splitting a sandwich with your hungry friend!
Therefore, $\angle ACD = \boxed{57^\circ}$.
Keep it fair and equal, my friend!
To find $\angle ACD$, we can start by drawing a diagram of the problem.
Step 1: Draw a triangle $ABC$ such that $AB$ and $AC$ are equal legs.
Here is a rough sketch of the triangle:
A
/ \
/ \
/________\
B C
Step 2: Draw point $D$ on $AB$ such that $CD = CB$.
Now, extend the line $CD$ so that it intersects $AB$ at point $D$ such that $CD = CB$.
A
/ \
/ \
/_____\
B C
.
.
D
Step 3: Measure the angle $\angle ADC$.
We are given that $\angle ADC = 114^\circ$.
Step 4: Find $\angle ACD$.
Since $CD = CB$, we have the isosceles triangle $CB = CD$. Therefore, $\angle CDB = \angle CBD$.
Since the sum of angles in a triangle is $180^\circ$, we can find $\angle CBD$:
$\angle CBD = (180^\circ - \angle CDB) / 2$
Since we are given $\angle ADC = 114^\circ$, we can substitute $\angle CDB = 180^\circ - \angle ADC$:
$\angle CBD = (180^\circ - 114^\circ) / 2$
$\angle CBD = 33^\circ$
Since $\angle CBD = \angle ACD$ in an isosceles triangle, we have $\angle ACD = 33^\circ$.
To find the measure of $\angle ACD$, we need to use the properties of isosceles triangles and the fact that angles in a triangle add up to $180^\circ$.
Let's start by drawing a diagram to visualize the situation:
```
C
/ \
/ \
/ \
/ \
A---------B
|
|
D
```
Since $AB$ and $AC$ are equal legs, we can mark the length of $AB$ and $AC$ as equal:
```
C
/ \
/ \
/ \
/ \
A---------B
|-------|
equal
```
Now, we know that $CD = CB$. Let's mark this on the diagram as well:
```
C
/ \
/ \
/ _\
/ _- \
A________B
|-------|
equal
```
We are given that $\angle ADC = 114^\circ$, so let's mark this angle in the diagram:
```
C
/ \
/ \
/ _\
/ _-\ |
A_______D
|------|
114°
```
To find $\angle ACD$, we need to use the fact that angles in a triangle add up to $180^\circ$. Since we have an isosceles triangle $ACD$, the base angles $\angle ACD$ and $\angle ADC$ are equal. Let's call the measure of $\angle ACD$ as $x$:
```
C
/ \
/ x \
/ _\
/ _-\ |
A_______D
|------|
114°
```
Since $ABCD$ is a quadrilateral, the sum of its interior angles is $360^\circ$. We can use this information to write an equation:
$\angle ACD + \angle ADC + \angle CDA + \angle CAB = 360^\circ$
Since $\angle ADC = 114^\circ$, we can substitute this value into the equation:
$\angle ACD + 114^\circ + \angle CDA + \angle CAB = 360^\circ$
Now, let's focus on the angles we know. We have $\angle ADC = 114^\circ$ and $\angle CDA = \angle CAB$ (because $AB$ and $AC$ are equal legs). So, we can rewrite the equation as:
$\angle ACD + 114^\circ + \angle CDA + \angle CDA = 360^\circ$
Simplifying the equation, we get:
$2\angle CDA + \angle ACD = 360^\circ - 114^\circ$
$2\angle CDA + \angle ACD = 246^\circ$
Now, recall that $\angle ACD = x$, so we can substitute it into the equation:
$2\angle CDA + x = 246^\circ$
Now we can solve for $x$, which represents the measure of angle $\angle ACD$:
$2\angle CDA = 246^\circ - x$
$\angle CDA = \frac{246^\circ - x}{2}$
Since $\angle CDA = \angle CAB$, we can substitute this into the equation:
$\angle CAB = \frac{246^\circ - x}{2}$
Since $\triangle ABC$ is an isosceles triangle, we know that $\angle CAB = \angle CBA$. So, we can substitute $\angle CBA$ for $\angle CAB$ in the equation:
$\angle CBA = \frac{246^\circ - x}{2}$
Now we have two angles: $\angle ACD = x$ and $\angle CBA = \frac{246^\circ - x}{2}$. Since these angles are equal, we can set up an equation:
$x = \frac{246^\circ - x}{2}$
Solving this equation, we get:
$2x = 246^\circ - x$
$3x = 246^\circ$
$x = \frac{246^\circ}{3}$
$x \approx 82^\circ$
Therefore, the measure of $\angle ACD$ is approximately $82^\circ$.