A parallel line through a point
Draw BC = 6 cm and AB = 4 cm with ∠ABC = 60°. Through A, draw a line parallel to BC.
Draw BC = 5.5 cm and AB = 4.5 cm with ∠ABC = 45°. Through A, draw a line parallel to BC.
Draw BC = 6 cm and AB = 4 cm with ∠ABC = 75°. Through A, draw a line parallel to BC.
Draw BC = 5 cm and AB = 4 cm with ∠ABC = 120°. Through A, draw a line parallel to BC.
1. Draw BC = 6 cm.
2. At B, construct ∠ABC = 60° and cut BA = 4 cm on its arm.
3. With B as the center and any radius, draw an arc that cuts BC at P and BA at Q.
4. With A as the center and the same radius, draw an arc on the other side of AB that cuts AB at R.
5. Open the compass to PQ. With R as the center and that radius, cut the arc at S.
6. Join SA and produce it both ways to T: ST ∥ BC, because ∠BAS = ∠ABC (alternate angles).
1. Draw BC = 5.5 cm.
2. At B, construct ∠ABC = 45° and cut BA = 4.5 cm on its arm.
3. With B as the center and any radius, draw an arc that cuts BC at P and BA at Q.
4. With A as the center and the same radius, draw an arc on the other side of AB that cuts AB at R.
5. Open the compass to PQ. With R as the center and that radius, cut the arc at S.
6. Join SA and produce it both ways to T: ST ∥ BC, because ∠BAS = ∠ABC (alternate angles).
1. Draw BC = 6 cm.
2. At B, construct ∠ABC = 75° and cut BA = 4 cm on its arm.
3. With B as the center and any radius, draw an arc that cuts BC at P and BA at Q.
4. With A as the center and the same radius, draw an arc on the other side of AB that cuts AB at R.
5. Open the compass to PQ. With R as the center and that radius, cut the arc at S.
6. Join SA and produce it both ways to T: ST ∥ BC, because ∠BAS = ∠ABC (alternate angles).
1. Draw BC = 5 cm.
2. At B, construct ∠ABC = 120° and cut BA = 4 cm on its arm.
3. With B as the center and any radius, draw an arc that cuts BC at P and BA at Q.
4. With A as the center and the same radius, draw an arc on the other side of AB that cuts AB at R.
5. Open the compass to PQ. With R as the center and that radius, cut the arc at S.
6. Join SA and produce it both ways to T: ST ∥ BC, because ∠BAS = ∠ABC (alternate angles).
Why it works
The two arcs have the same radius and the chord RS equals PQ, so the angle at A is a copy of the angle at B: ∠BAS = ∠ABC. They are alternate angles between ST and BC with AB crossing both, and equal alternate angles make the lines parallel.