Question

Через вершину В трикутника АВС до його площини проведено перпендикуляр МВ. Пряма , яка проходить через точку М , перпендикулярна выдрызку АС і перетинає цей відрізок в його середині. Довести що трикутник АВС рівнобедренний. Через вершину В треугольника АВС к его плоскости проведен перпендикуляр МВ. Прямая, проходящая через точку М, перпендикулярная выдрызку АС и пересекает этот отрезок в его середине. Доказать что треугольник АВС ривнобедренний

Answer 1

Накресливши пряму ВМ, яка проходить через середину сторони АС ми отримали два прямокутних трикутника АВМ і СВМ. ВМ - їх спільний катет, а оскільки перпендикуляр ВМ проведений до середина АС, то АМ=АС. Отже, трикутники АВМ і СВМ будуть рівними за 3 сторонами. А це означає, що АВ=ВС, що й треба було довести

Answer 2

I can answer your question in detail. You have asked me to prove that the triangle ABC is isosceles, given that a perpendicular MV is drawn from the vertex B to the plane of the triangle, and a line passing through the point M is perpendicular to the segment AC and intersects it at its midpoint. Here is my solution:

Let D be the midpoint of AC. Then, by the definition of a perpendicular, we have that MD is perpendicular to AC. Since MD is also perpendicular to MV, we have that MD is the altitude of the triangle MBC. Therefore, by the theorem of the right triangle altitude, we have that MD divides the triangle MBC into two triangles that are similar to the original triangle MBC and to each other. In particular, we have that the triangles MBD and MDC are similar.

Now, since D is the midpoint of AC, we have that AD = DC. By the corresponding sides of similar triangles, we have that BD = MC. Therefore, by the side-angle-side criterion of triangle congruence, we have that the triangles ABD and BMC are congruent. Hence, we have that the angles ABD and BMC are equal.

Finally, by the angle addition postulate, we have that the angle ABC is equal to the sum of the angles ABD and BMC. Since these two angles are equal, we have that the angle ABC is bisected by the segment BD. Therefore, by the angle bisector theorem, we have that AB = BC. This means that the triangle ABC is isosceles, as required.

I hope this answer helps you understand the problem. If you want to learn more about triangles and their properties, you can check out some of the web search results that I found for you . Have a nice day!