angle bisectors: solve for x and y.
4. given: $\overline{at}$ bisects $\angle cab$
5. given: $\overline{ba}$ bisects $\angle rbt$
6. given: $\overline{ba}$ bisects $\angle cad$
segment bisectors. solve for x.
7. given: n is a midpoint.
8. given: $\overline{le}$ bisects $\overline{na}$
9. given: $\overline{ab}$ is a perpendicular bisector.

Question

Accepted Answer

### Problem 4:
# Explanation:
## Step1: Use Angle Bisector Definition
Since $ \overline{AT} $ bisects $ \angle CAB $, $ \angle CAT = \angle TAB = 52^\circ $, so $ x = 52 $.
## Step2: Use Full Angle Sum
A full angle at $ A $ is $ 360^\circ $. So $ y + 52 + 52 = 360 $.
$$ y + 104 = 360 $$
## Step3: Solve for $ y $
Subtract 104 from both sides:
$$ y = 360 - 104 = 256 $$
# Answer:
$ x = 52 $, $ y = 256 $

### Problem 5:
# Explanation:
## Step1: Use Angle Bisector Definition
Since $ \overline{BA} $ bisects $ \angle RBT $, $ \angle RBA = \angle ABT = 26^\circ $, so $ x = 26 $.
## Step2: Use Triangle Angle Sum (if needed, but here $ y $ in triangle: $ 180 - 26 - 26 - \angle T $, but from diagram, likely $ y = 180 - 26 - 26 = 128 $? Wait, no, the diagram shows $ y $ as the angle at $ A $. Wait, in triangle $ RBA $, if $ \angle RBA = 26^\circ $, $ \angle R = 90^\circ $ (maybe isoceles), so $ y = 180 - 26 - 26 = 128 $? But the given $ x = 26 $, so $ x = 26 $, and $ y = 180 - 26 - 26 = 128 $ (assuming triangle angle sum).
# Answer:
$ x = 26 $, $ y = 128 $ (assuming triangle angle sum)

### Problem 6:
# Explanation:
## Step1: Use Angle Bisector and Linear Pair
$ \angle CAD $ and $ 108^\circ $ are linear pair? Wait, $ \overline{BA} $ bisects $ \angle CAD $, so $ \angle CAB = \angle BAD = x $. The straight line at $ A $: $ x + x + 108 = 180 $ (linear pair sum to $ 180^\circ $)? Wait, no, the diagram: $ \angle CAD $ is bisected by $ BA $, so $ \angle CAB = \angle BAD = x $, and $ \angle CAD + 108^\circ = 180^\circ $ (linear pair). So $ 2x + 108 = 180 $.
## Step2: Solve for $ x $
$$ 2x = 180 - 108 = 72 $$
$$ x = 36 $$
## Step3: Use Triangle Angle Sum (triangle $ BAD $ is isoceles, so $ y = 180 - 36 - 36 = 108 $? Wait, no, $ \angle BAD = x = 36 $, $ \angle ADB = 36 $ (isoceles), so $ y = 180 - 36 - 36 = 108 $? Wait, no, the angle at $ B $ is $ y $, so $ y = 180 - 36 - 36 = 108 $? Wait, maybe:
Linear pair: $ \angle CAD + 108^\circ = 180^\circ $ ⇒ $ \angle CAD = 72^\circ $. Bisected by $ BA $, so $ \angle CAB = \angle BAD = 36^\circ $ (so $ x = 36 $). Then in triangle $ BAD $, $ \angle BAD = 36^\circ $, $ \angle ADB = 36^\circ $ (isoceles), so $ y = 180 - 36 - 36 = 108^\circ $.
# Answer:
$ x = 36 $, $ y = 108 $

### Problem 7:
# Explanation:
## Step1: Use Midpoint Definition
Since $ N $ is midpoint, $ MN = NP $. So $ 9x - 1 = 26 $.
## Step2: Solve for $ x $
$$ 9x = 26 + 1 = 27 $$
$$ x = 3 $$
# Answer:
$ x = 3 $

### Problem 8:
# Explanation:
## Step1: Use Segment Bisector Definition
Since $ \overline{LE} $ bisects $ \overline{NA} $, the segments are equal. So $ x + 41 = 3x + 17 $ (wait, no: $ LE $ bisects $ NA $, so the intersection point divides $ NA $ into two equal parts. Wait, $ NA $ is a segment, bisected by $ LE $, so $ 4x = x + 41 $? Wait, no, the diagram: $ N $ to $ A $ is $ 4x $, and $ L $ to $ E $ intersects $ NA $ at midpoint, so $ x + 41 = 3x + 17 $ (since $ LE $ is bisected? Wait, no, $ LE $ bisects $ NA $, so the midpoint of $ NA $ is on $ LE $, so $ x + 41 = 3x + 17 $.
## Step2: Solve for $ x $
$$ 41 - 17 = 3x - x $$
$$ 24 = 2x $$
$$ x = 12 $$
# Answer:
$ x = 12 $

### Problem 9:
# Explanation:
## Step1: Use Perpendicular Bisector Definition
A perpendicular bisector means the segments are equal and angles are $ 90^\circ $, but here, the segments $ 8x - 26 $ and $ 5x + 7 $ are equal (since $ AB $ is perpendicular bisector, so the two segments are equal). So $ 8x - 26 = 5x + 7 $.
## Step2: Solve for $ x $
$$ 8x - 5x = 7 + 26 $$
$$ 3x = 33 $$
$$ x = 11 $$
Check: $ 8(11) - 26 = 88 - 26 = 62 $, $ 5(11) + 7 = 55 + 7 = 62 $. Also, $ 6x + 1 = 67 $, but since it's perpendicular bisector, the other segment is equal too (but here we solve $ 8x - 26 = 5x + 7 $).
# Answer:
$ x = 11 $

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QUESTION IMAGE

Question

Explanation:

Problem 4:

Step1: Use Angle Bisector Definition

Step2: Use Full Angle Sum

Step3: Solve for \( y \)

Step1: Use Angle Bisector Definition

Step1: Use Angle Bisector and Linear Pair

Step2: Solve for \( x \)

Step3: Use Triangle Angle Sum (triangle \( BAD \) is isoceles, so \( y = 180 - 36 - 36 = 108 \)? Wait, no, \( \angle BAD = x = 36 \), \( \angle ADB = 36 \) (isoceles), so \( y = 180 - 36 - 36 = 108 \)? Wait, no, the angle at \( B \) is \( y \), so \( y = 180 - 36 - 36 = 108 \)? Wait, maybe:

Answer:

Problem 5: