Question

ДАЮ 35 БАЛОВ У трикутнику abc проведена бісектриса bd . Довести що якшо кут bdc тупий то середина сторони ac належить променю ds .

Answer 1

Готово.....................................

Answer 2

Given Information:

We are given a triangle ABC and a bisector BD is drawn. We need to prove that if angle BDC is obtuse, then the midpoint of side AC lies on the ray DS.

Proof:

To prove this statement, we will use the properties of angles and triangles.

Let's assume that angle BDC is obtuse. This means that the measure of angle BDC is greater than 90 degrees.

Now, let's consider the triangle ABC. Since BD is the bisector of angle B, it divides angle B into two equal angles, angle ABD and angle CBD.

Since angle BDC is obtuse, angle CBD must be acute. This is because the sum of the measures of angles in a triangle is always 180 degrees, and if angle BDC is obtuse, then the sum of angles ABD and CBD must be less than 90 degrees.

Now, let's consider the triangle BDC. Since angle CBD is acute, angle BCD must be obtuse. This is because the sum of the measures of angles in a triangle is always 180 degrees, and if angle CBD is acute, then the sum of angles BCD and BDC must be greater than 90 degrees.

Since angle BCD is obtuse, the side AC must intersect the extension of side BD beyond point D. Let's call the intersection point E.

Now, let's consider the triangle AEC. Since E lies on the extension of side BD, which is the bisector of angle B, angle AED and angle CED must be equal. This is because the bisector of an angle divides the opposite side into two segments that are proportional to the adjacent sides of the angle.

Since angle AED and angle CED are equal, the triangle AEC must be isosceles. In an isosceles triangle, the perpendicular bisector of the base passes through the vertex angle.

Therefore, the midpoint of side AC lies on the ray DS, which is the extension of side BD beyond point D.

In conclusion, if angle BDC is obtuse, then the midpoint of side AC lies on the ray DS.