Question

how to prove that the corresponding angle formed by transversal of two parallel lines are of equal measures

Answer 1

Suppose you have two parallel lines cut by a transversal.

Due to the straight angle (linear pair) theorem, we know that

{m∠2+m∠3=180˚m∠5+m∠6=180˚

Thought the transitive property, we can say that

m∠2+m∠3=m∠5+m∠6×× (1)

Though the alternate interior angles theorem, we know that

m∠3=m∠5

Use substitution in (1):

m∠2+m∠3=m∠3+m∠6

Subtract m∠3 from both sides of the equation

m∠2=m∠6

∴∠2≅∠6

Thus ∠2 and ∠6 are corresponding angles and have proven to be congruent.

Answer 2

Answer:

∠2 and ∠6 are corresponding angles and are equal.

Step-by-step explanation:

Consider two parallel lines p and q cut by a transversal T.

In total, eight angles are formed, 4 at each point.

The angles that occupy the same relative position at each intersection point of parallel lines with a transversal are called corresponding angles.

In the figure, ∠1 & ∠5, ∠2 & ∠6, ∠3 & ∠7 and ∠4 & ∠8 form the pairs of corresponding angles.  

Consider the straight angles formed by ∠2 & ∠3 and ∠5 & ∠6.

Since the pairs form a linear pair, the sum of the angles is equal to 180°.

Therefore $\angle 2+\angle 3=180^o$ and $\angle 5+\angle 6=180^o$

Using the transitive property of equality, we get

$\angle 2+\angle 3=\angle 5+\angle 6$                                        ...(1)

The ∠3 and ∠5 are the alternate interior angles, therefore they are equal.

Thus $\angle 3=\angle 5$

Substituting this in the equation (1)

$\angle 2+\angle 5=\angle 5+\angle 6$

Subtracting measure of ∠5 from both sides of the equation, we get

$\angle 2=\angle 6$

Thus, ∠2 and ∠6 are corresponding angles and are equal.

Hence proved.

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