The problem shows a triangle $\triangle ABC$ with a line $ED \parallel BC$, and angles and expressions provided at various points.

Step 1: Solve for $x$ (18.1)

Since $ED \parallel BC$, we can use the concept of corresponding angles in parallel lines.

1. Angle $EAD = 3x - 15^\circ$
2. Angle $D_1 = 4x + 20^\circ$
3. Angle $A = 70^\circ$

In a triangle, the sum of the angles equals $180^\circ$. Therefore, for $\triangle AED$: $\angle A + \angle EAD + \angle D_1 = 180^\circ$ Substitute the given angles: $70^\circ + (3x - 15^\circ) + (4x + 20^\circ) = 180^\circ$ Simplify: $70^\circ + 3x - 15^\circ + 4x + 20^\circ = 180^\circ$ $70^\circ - 15^\circ + 20^\circ + 7x = 180^\circ$ $75^\circ + 7x = 180^\circ$ Solve for $x$: $7x = 180^\circ - 75^\circ$ $7x = 105^\circ$ $x = \frac{105^\circ}{7}$ $x = 15^\circ$

Step 2: Calculate the size of $\hat{D}_1$ (18.2)

Now that $x = 15^\circ$, substitute into the expression for $\hat{D}_1 = 4x + 20^\circ$: $\hat{D}_1 = 4(15^\circ) + 20^\circ = 60^\circ + 20^\circ = 80^\circ$

Final Answers:

• 18.1: $x = 15^\circ$
• 18.2: $\hat{D}_1 = 80^\circ$

Would you like a detailed explanation of the angle relationships?

Here are 5 questions for further practice:

1. How can you prove the sum of angles in any triangle equals $180^\circ$?
2. What are corresponding angles, and why are they equal when lines are parallel?
3. How do you solve for unknown angles in a quadrilateral?
4. What is the difference between alternate and corresponding angles?
5. Can the angles formed in a triangle ever exceed 180 degrees?

Tip: Always check the parallel lines and corresponding angles when dealing with transversals in triangles!