Question

 В равнобедренном треугольнике АВС с основанием АС проведена медиана ВD. Точка Р -  середина стороны ВА, точка  К – середина стороны  ВС.  Докажите равенство треугольников ВDР и ВDК. Изобразить чертеж. ​

Answer 1

Ответ:

ΔАВС ,  АВ=ВС  ,  АР=РВ ,  ВК=КС  .

 Так как АВ=ВС , то и половины этих сторон равны между собой: АР=РВ=ВК=КС .

Рассм. ΔВДР и ΔВДС . У них ВР=ВК , ВД - общая сторона и ∠РВД=∠КВД , так как медиана равнобедр. тр-ка , проведённая к основанию, является ещё и биссектрисой .

Значит, ΔВДР=ΔВДС по 1 признаку равенства треугольников .

Answer 2

Here is my answer to your question:

To prove that the triangles BDR and BDK are equal, we need to show that they have two sides and the included angle equal. We can use the following facts:

- The triangle ABC is isosceles, so AB = BC and angle A = angle C. - The point D is the midpoint of AC, so AD = DC and BD is the median of ABC. - The points P and K are the midpoints of AB and BC, respectively, so AP = PB, BK = KC, and PK is parallel to AC.

Using these facts, we can see that:

- BD is a common side for both triangles BDR and BDK. - Angle BDR = angle BDK, because they are alternate interior angles formed by the transversal BD and the parallel lines AC and PK. - BR = BK, because BR = PB + PR, BK = KC + CK, and PB = KC, PR = CK, by the midpoint theorem.

Therefore, by the SAS criterion, the triangles BDR and BDK are equal.

Here is a diagram of the situation:

    A
   / \
  /   \
 P-----D-----K
/ \   / \   / \
/   \ /   \ /   \
B----R-----C----S

I hope this helps you understand the problem.