Отрезки AB и CD пересекаются в их общей середине а) доказать что угол BCD= углу ADC б) найти BC, если AD=6 см

To prove that angle BCD is equal to angle ADC, we can use the fact that AB and CD intersect at their common midpoint.

Since AB and CD intersect at their common midpoint, we can draw two lines from the midpoint to points B and D. Let's call the point of intersection of the line through the midpoint and point B as E, and the point of intersection of the line through the midpoint and point D as F.

Now, we have two triangles AED and CFD. Both triangles share the side AD, and we know that the segments AB and CD intersect at the midpoint, so AE = EF = FB and CF = FD = DE.

Since AE = EF = FB, triangle AEB is an isosceles triangle. Similarly, CF = FD = DE, triangle CFD is also an isosceles triangle.

Now, let's consider angle BCD. Since triangle CFD is isosceles, angle CFD is equal to angle CDF. The same goes for triangle AEB, where angle AEB is equal to angle ABE.

Since angle CDF is equal to angle ABE (as they are vertical angles), we can conclude that angle BCD is equal to angle ADC.

Now, to find BC, we need to use the fact that AD is equal to 6 cm. Since AD is the base of both triangles AED and CFD, and they are similar by angle-angle similarity, we can set up the following proportion:

BC/CF = AD/DE

Substituting the given value, we have:

BC/CF = 6/DE

Since we know that CF = FD = DE, we can simplify the proportion:

BC/DF = 6/DE

Since DF = DE, we can further simplify:

BC = 6

Therefore, BC is equal to 6 cm.