0
$\begingroup$

I need to prove Tranversal parallel lines theorem that says: If two parallel lines are cut by a transversal, the corresponding angles are congruent, the alternate angles are congruent, and the consecutive angles are supplementary. And other way around, If two lines are cut by a transversal and the corresponding angles are congruent, then the lines are parallel. Same goes if the alternate angles are congruent, and if the consecutive angles are supplementary.

This what I have: Two parallel lines $p$ and $q$ are cut by tranversal $t$. Becuase $p$ and $q$ are parallel, there's no intersection point between them. There's also one proposition that says: "If two different lines close with some other third line congruent angles then that two lines don't intersect". And because of that the corresponding angles must be congruent.

$\endgroup$
0
$\begingroup$

Euclid's fifth postulate says, "If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough." So when parallel lines $p$ and $q$ are crossed by $t$, the sum of the angles on one side cannot be less than $180$ degrees, and it cannot be greater, too, as that would mean the sum of the angles on the other side would be less than $180$ degrees. So it must be equal to $180$ degrees, and the rest follows.

The converse can be built on the same lines.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.