Question

на диагонали АС параллелограмма АВСД отметили точки Р и К так, что АР=СК.Докажите, что угол АДР =углу СВБ

Answer 1

Здесь необходимо знать 1-й признак равенства треугольников. и да, вот еще что: старайся правильно и корректно писать условия задачи, ведь это может привести к ошибке.

Answer 2

I will try to answer your question in detail. You asked me to prove that the angle ADR is equal to the angle CBK in a parallelogram ABCD, where points P and K are marked on the diagonal AC such that AP = CK (point P lies between points A and K).

To prove this, we can use the following steps:

1. Since ABCD is a parallelogram, we have AB || CD and AD || BC by definition. 2. Since AP = CK, we have AK = AC - AP = AC - CK = AK by subtraction. Therefore, AK is the midpoint of AC by definition. 3. Since AK is the midpoint of AC, we have AK || BD by the midpoint theorem. 4. Since AK || BD and AB || CD, we have AKCD is a parallelogram by the converse of the definition of a parallelogram. 5. Since AKCD is a parallelogram, we have ADK = CKD by the opposite angles theorem. 6. Since ADK = CKD and AD || BC, we have ADP = CBK by the alternate interior angles theorem.

Therefore, we have proved that ADP = CBK as required.

You can also see a video explanation of this proof [here](https://www.youtube.com/watch?v=aoSqQIFPqUE) or read a similar problem [here](https://uchi.ru/otvety/questions/na-diagonali-as-parallelogramma-abcd-otmetili-tochki-r-i-k-tak-chto-ar-sk-tochka-r-lezhit). I hope this helps you understand the solution.