Question

Отрезки АВ и CD пересекаются в точке О,являющейся их серединой.Докажите параллельность прямых АС и BD?С решением пожалуста!

Answer 1

По условию AO=BO,OC=OD

Углы AOC и BOD равны, как вертикальные.

 

Треугольники AOC и BOD равны за двумя сторонами и углом между ними (AO=OB,CO=OD, углы AOC и BOD равны)

 

с равенства треугольников имеем равенство углов

угол OAC= угол OBD, иначе

угол BAC=угол ABD

углы BAC и ABD внутренние разносторонние при секущей AB, поэтому

прямые AC и BD параллельны по признаку паралельности прямых. Доказано

Answer 2

Proof of Parallel Lines AC and BD

To prove that lines AC and BD are parallel, we can use the given information that segments AB and CD intersect at point O, which is their midpoint.

Let's assume that lines AC and BD are not parallel. In that case, they would intersect at some point E.

Now, let's consider triangle AEO. Since O is the midpoint of segment AB, we can conclude that AO is congruent to BO. Similarly, since O is the midpoint of segment CD, CO is congruent to DO.

Using the transitive property of equality, we can say that AO is congruent to BO, and CO is congruent to DO.

Now, let's consider triangle CEO. Since CO is congruent to DO, and CE is a common side, we have two sides of triangle CEO congruent to two sides of triangle AEO.

According to the Side-Side-Side (SSS) congruence criterion, if two sides and the included angle of one triangle are congruent to the corresponding sides and included angle of another triangle, then the two triangles are congruent.

Therefore, triangle CEO is congruent to triangle AEO.

As a result, the corresponding angles of congruent triangles CEO and AEO are congruent.

Let's denote the angle at point E as angle CED and the angle at point A as angle AEO. Since triangle CEO is congruent to triangle AEO, angle CED is congruent to angle AEO.

Now, let's consider triangle BEO. Since O is the midpoint of segment AB, we can conclude that AO is congruent to BO.

Using the transitive property of equality, we can say that AO is congruent to BO.

Now, let's consider triangle BEO. Since AO is congruent to BO, and BE is a common side, we have two sides of triangle BEO congruent to two sides of triangle AEO.

According to the Side-Side-Side (SSS) congruence criterion, if two sides and the included angle of one triangle are congruent to the corresponding sides and included angle of another triangle, then the two triangles are congruent.

Therefore, triangle BEO is congruent to triangle AEO.

As a result, the corresponding angles of congruent triangles BEO and AEO are congruent.

Let's denote the angle at point E as angle BED and the angle at point A as angle AEO. Since triangle BEO is congruent to triangle AEO, angle BED is congruent to angle AEO.

Now, we have angle CED congruent to angle AEO and angle BED congruent to angle AEO.

According to the Transitive Property of Congruence, if two angles are congruent to the same angle, then they are congruent to each other.

Therefore, angle CED is congruent to angle BED.

However, this is a contradiction because angles CED and BED are corresponding angles formed by the transversal line AC intersecting the lines CD and AB.

According to the Converse of the Corresponding Angles Postulate, if two lines are cut by a transversal and corresponding angles are congruent, then the lines are parallel.

Since we have reached a contradiction, our assumption that lines AC and BD are not parallel must be false.

Therefore, lines AC and BD are parallel.