If two non parallel lines are perpendicular to other two non parallel lines prove that acute angel between first two non parallel lines is equal to acute angle between other two non parallel lines

To prove that the acute angle between the first two non-parallel lines is equal to the acute angle between the other two non-parallel lines, we can use the concept of perpendicular lines and properties of parallel lines.

Let's assume we have four lines: line AB and line CD are perpendicular to each other and line AC and line BD are also perpendicular to each other. These lines are non-parallel.

To prove that the acute angle between AB and AC is equal to the acute angle between CD and BD, we can make use of the properties of parallel lines.

Construct a line EF such that EF is parallel to AB and CD, as shown:

E-----------------F
/ \
/ \
A-------C B-------D
\ /
\ /
G-----------------H

Since EF is parallel to AB and CD, the alternate interior angles EAG and FCH are equal, as well as the alternate interior angles EBF and FDG.

Now, consider the triangles EAG and FCH. We know that two of their angles are equal (the right angle formed by lines AB and CD), and we have just shown that the alternate interior angles EAG and FCH are equal. Therefore, the third angle of each triangle, the acute angle formed by lines AG and HC, must also be equal.

Similarly, for the triangles EBF and FDG, since the alternate interior angles EBF and FDG are equal, the third angle formed by lines BG and HD must also be equal.

Therefore, we have proven that the acute angle between AB and AC is equal to the acute angle between CD and BD.

To prove that the acute angle between the first two non-parallel lines is equal to the acute angle between the other two non-parallel lines, we can use the concept of perpendicular lines.

Let's consider the given lines as follows:

Line 1: AB
Line 2: CD
Line 3: EF
Line 4: GH

We are given that Line 1 (AB) is perpendicular to Line 3 (EF) and Line 2 (CD) is perpendicular to Line 4 (GH).

To prove that the acute angle between Line 1 and Line 2 is equal to the acute angle between Line 3 and Line 4, we can use the fact that perpendicular lines form right angles.

1. Since Line 1 (AB) is perpendicular to Line 3 (EF), we have ∠ABE = 90° and ∠DEF = 90°.

2. Similarly, Line 2 (CD) being perpendicular to Line 4 (GH), we have ∠CDF = 90° and ∠GHF = 90°.

3. Now, we can define two acute angles: ∠BAE between Line 1 (AB) and Line 3 (EF), and ∠DCG between Line 2 (CD) and Line 4 (GH).

4. Since the sum of angles in a triangle is 180°, we can express ∠BAE and ∠DCG in terms of the given right angles:

∠BAE = 180° - ∠ABE - ∠DEF
∠DCG = 180° - ∠CDF - ∠GHF

5. Plugging in the known right angles (∠ABE = 90°, ∠DEF = 90°, ∠CDF = 90°, ∠GHF = 90°), we get:

∠BAE = 180° - 90° - 90° = 0°
∠DCG = 180° - 90° - 90° = 0°

6. Since both ∠BAE and ∠DCG equal 0°, it implies that the acute angle between Line 1 (AB) and Line 3 (EF) is equal to the acute angle between Line 2 (CD) and Line 4 (GH).

Thus, we have proven that the acute angle between the first two non-parallel lines is equal to the acute angle between the other two non-parallel lines.