Question

Дано:треугольник ABC и треугольник DEF,AB=BC;DE=FE.угол 1 равен углу 2.Доказать AB II EF.Срочно

Answer 1

Есть рисунок к задачи?

Answer 2

Given Information:

We are given two triangles, triangle ABC and triangle DEF, with the following properties: - AB = BC - DE = FE - Angle 1 is equal to angle 2.

We need to prove that AB is parallel to EF.

Proof:

To prove that AB is parallel to EF, we can make use of the fact that corresponding angles of two parallel lines are equal.

Let's assume that AB and EF are not parallel. In that case, there must be a point G on AB and a point H on EF such that GH is a transversal intersecting AB and EF.

Since AB = BC, we can draw a line segment CD such that CD is parallel to AB and passes through point H. This is possible because any line can be drawn parallel to another line through a given point.

Now, we have two parallel lines AB and CD, and a transversal GH intersecting them. According to the properties of parallel lines, the corresponding angles formed by the transversal GH and the parallel lines AB and CD should be equal.

Let's consider angle 1 and angle 2. Since angle 1 is equal to angle 2, the corresponding angles formed by the transversal GH and the parallel lines AB and CD are also equal.

Therefore, angle 1 is equal to angle 2 is equal to angle 3 (the corresponding angle to angle 1).

Now, let's consider triangle ABC and triangle CDH. We have: - AB = CD (by construction) - BC = DH (by construction) - Angle 1 = Angle 3 (corresponding angles)

By the Side-Angle-Side (SAS) congruence criterion, we can conclude that triangle ABC is congruent to triangle CDH.

Since corresponding sides of congruent triangles are equal, we have AB = CD. But we already know that AB = BC, so this implies BC = CD.

However, this contradicts the fact that BC is parallel to CD. If BC = CD, then BC and CD would coincide, and they cannot be parallel.

Therefore, our assumption that AB is not parallel to EF is incorrect. Hence, we can conclude that AB is parallel to EF.

Therefore, AB II EF.

Please note that the proof provided above is based on the given information and the properties of parallel lines and congruent triangles.