Question

" На сторонах AB і BC трикутника ABC позначили від- повідно точки D i Е так, що ZEAC = ZDCA. Відрізки АЕ і CD перетинаються в точці F, DF = EF. Доведіть, що трикутник ABC рівнобедрений.​

Answer 1

вот ответ на данный вопрос по фотке

Answer 2

Triangle ABC and Points D, E, and F

To answer the question, we need to prove that triangle ABC is isosceles, or in other words, that AB = AC.

Given that ZEAC = ZDCA, we can infer that triangles ZEA and ZDC are similar by the Angle-Angle (AA) similarity criterion. This means that the corresponding angles of these triangles are equal.

Since ZEAC = ZDCA, we have:

ZEA = ZDC [[1]]

Now, let's consider triangles AEF and CDF. We know that DF = EF, and we want to prove that AE = CD.

Since triangles ZEA and ZDC are similar, we can conclude that triangles AEF and CDF are also similar by the Angle-Angle (AA) similarity criterion. This means that the corresponding angles of these triangles are equal.

Since ZEA = ZDC, we have:

AEF = CDF [[2]]

Now, let's consider triangles AEF and ABC. We know that AEF = CDF and DF = EF, and we want to prove that AE = AB.

Since triangles AEF and CDF are similar, we can conclude that triangles AEF and ABC are also similar by the Angle-Angle (AA) similarity criterion. This means that the corresponding angles of these triangles are equal.

Since AEF = CDF, we have:

AE = AB [[3]]

Therefore, we have proved that triangle ABC is isosceles, or in other words, that AB = AC.