Let’s solve the problems on the image step by step:

Problem 3: Find $m \angle BAD$

Given:

• $AC$ is an altitude.
• $\triangle BAD$ has two expressions for side lengths $4x + 4$ and $6x - 6$.
• $\angle A = 38^\circ$.

Solution:

1. The altitude $AC$ makes $\triangle BAC$ and $\triangle CAD$ right triangles.
2. The triangle $\triangle BAD$ is isosceles since $AC$ is an altitude and bisects the base $BD$.
3. Use the property of isosceles triangles: $\angle B = \angle D$.
4. The sum of angles in a triangle is $180^\circ$: $\angle BAD + \angle B + \angle D = 180^\circ$ Substituting $\angle B = \angle D$, and given $\angle A = 38^\circ$: $38^\circ + 2\angle B = 180^\circ$ $2\angle B = 142^\circ$ $\angle B = \angle D = 71^\circ$ Hence: $m \angle BAD = 38^\circ$

Problem 4: Find missing angles in $\triangle GEM$

Given:

• $AE$ is a perpendicular bisector.
• $AE$ divides $\triangle GEM$ into two congruent right triangles.
• $\angle E = 90^\circ$.

Solution:

1. Since $AE$ is a perpendicular bisector, $\triangle GEM$ is isosceles ($GE = GM$).
2. The angles opposite equal sides are equal: $\angle G = \angle M$
3. The sum of the angles in a triangle is $180^\circ$: $\angle G + \angle M + \angle E = 180^\circ$ Substituting $\angle E = 90^\circ$: $\angle G + \angle M = 90^\circ$ Since $\angle G = \angle M$: $2\angle G = 90^\circ$ $\angle G = \angle M = 45^\circ$ Thus, the missing angles are: $m\angle G = 45^\circ, \quad m\angle M = 45^\circ, \quad m\angle E = 90^\circ$

Problem 5: Find missing side lengths ($\triangle USG$)

Given:

• $BS$ is a perpendicular bisector.
• $BU = 13$, $BG = 5$.

Solution:

1. $BS$ being a perpendicular bisector implies $SU = SG$.
2. Use the Pythagorean theorem in $\triangle BUS$ (right triangle): $BS^2 + SU^2 = BU^2$ Substituting $BU = 13$: $BS^2 + SU^2 = 169$
3. Use the Pythagorean theorem in $\triangle BSG$ (right triangle): $BS^2 + SG^2 = BG^2$ Substituting $BG = 5$: $BS^2 + SG^2 = 25$
4. Since $SU = SG$, substitute $SG = SU$: Solve the system of equations: $BS^2 + SU^2 = 169$ $BS^2 + SU^2 = 25$
5. Combine to find $BS$ and $SU$.

Would you like further clarification or details on this? Let me know!

Follow-up Questions:

1. What is the significance of a perpendicular bisector in triangle geometry?
2. How does the property of isosceles triangles simplify calculations in problem 4?
3. Why do we rely on the Pythagorean theorem in problem 5?
4. Can you generalize how altitudes affect the angles and sides in triangles?
5. How would the solutions change if additional information about side lengths were given?

Tip: Always label congruent parts of a triangle when working with bisectors to avoid confusion.