Question

Точка Е принадлежит большему основанию [АВ] трапеции АВСD так, что [АЕ] конгруэнтен [DC].Докажите, что отрезки АС и DE делятся пополам точкой пересечения.

Answer 1

           Д                  С

                  О

А                     Е                      В

ДС=АЕ и   ДСIIАЕ (основания трапеции парал., точка Е принадледит АВ), то  и СЕIIАД.  АДСЕ-параллелограмм, а у него диагонали точкой пересечения делятся пополам. Значит АО=ОС и ДО=ОЕ

Или доказываем равенство тр-в АОЕ и ДОС по стороне АЕ=ДС и прилежащим к ним углам (внутренние накрест лежащие).

Answer 2

Given Information:

We are given that point E belongs to the larger base [AB] of trapezoid ABCD, such that segment [AE] is congruent to [DC]. We need to prove that segments AC and DE bisect each other at the point of intersection.

Proof:

To prove that segments AC and DE bisect each other at the point of intersection, we can use the concept of similar triangles and the properties of trapezoids.

1. Let's consider triangle AED and triangle CDE. Since segment [AE] is congruent to [DC], we can conclude that triangle AED is congruent to triangle CDE by the Side-Side-Side (SSS) congruence criterion.

2. Since triangle AED is congruent to triangle CDE, their corresponding angles are congruent as well. Therefore, angle AED is congruent to angle CDE.

3. Now, let's consider trapezoid ABCD. Since point E belongs to the larger base [AB], we can conclude that angle AED is congruent to angle CDE by the Alternate Interior Angles Theorem.

4. Since angle AED is congruent to angle CDE and angle AED is congruent to angle CDE, we can conclude that angle AEC is congruent to angle DEC by the Transitive Property of Congruence.

5. By the Angle Bisector Theorem, we know that if a line bisects an angle of a triangle, it divides the opposite side into segments that are proportional to the lengths of the other two sides. In this case, segment AC bisects angle AEC, so it divides segment DE into two equal halves.

6. Therefore, we can conclude that segments AC and DE divide each other in half at the point of intersection.

Thus, we have proved that segments AC and DE divide each other in half at the point of intersection.

Note: The given information and proof are based on the provided question and my understanding of the topic. If you need further clarification or have additional information, please let me know.