Question

Дано: ΔABC, AA1, BB1 - биссектрисы. AA1∩BB1=0. ∠ABC=30, ∠AOB=107°. Доказать: ΔABC не является остроугольным.С ПОДРОБНЫМ ОБЪЯСНЕНИЕМ!

Answer 1

Рассмотрим  Δ АОВ. ∠AOB=107° - по условию. Так как ВВ1 биссектиса и делит угол АВС пополам     ∠ АВО= 1/2∠АВС=1/2*30=15°

Сумма углов  треугольника  равна 180°. ∠ ОАВ=180-107-15=58°

Рассмотрим  Δ АВС . Так как АА 1 биссектрисса  и делит  угол  САВ пополам, ∠ САВ = 2*∠ОАВ=2*58=116°

∠ АСВ = 180-116-30= 34°

Остроугольный треугольник - это треугольник, в котором все три угла острые, т.е. меньше 90°.

В ΔABC  два острых угла  ∠ АВС=30 °, ∠АСВ=34 °, а ∠САВ=116 °. Значит ΔАВС не острый, а тупоугольный.

 

Answer 2

Given Information:

We are given the following information: - ΔABC is a triangle. - AA1 and BB1 are the bisectors of angles A and B, respectively. - AA1 intersects BB1 at point O and the intersection point O is the origin (0,0) on the coordinate plane. - ∠ABC = 30°. - ∠AOB = 107°.

Proof that ΔABC is not acute-angled:

To prove that ΔABC is not acute-angled, we need to show that at least one angle in the triangle is not less than 90°.

Let's assume that ΔABC is acute-angled. In an acute-angled triangle, all angles are less than 90°. Therefore, we need to show that at least one angle in ΔABC is not less than 90°.

Since AA1 and BB1 are bisectors, they divide angles A and B into two equal parts. Let's denote the intersection point of AA1 and BB1 as O.

From the given information, we know that ∠ABC = 30° and ∠AOB = 107°.

To find the measure of angle A, we can use the fact that AA1 is the bisector of angle A. Since AA1 intersects BB1 at point O, we can use the angle bisector theorem to find the measure of angle A.

According to the angle bisector theorem, the ratio of the lengths of the segments formed by the intersection point on the bisector is equal to the ratio of the lengths of the corresponding sides of the triangle.

Let's denote the length of segment AO as x and the length of segment BO as y. Then, we have:

AO/OB = AB/BA1

Since point O is the origin (0,0) on the coordinate plane, the coordinates of points A and B can be denoted as (x,0) and (0,y), respectively.

Using the distance formula, we can find the lengths of AB and BA1:

AB = √((x-0)^2 + (0-0)^2) = √(x^2) = x BA1 = √((0-x)^2 + (y-0)^2) = √(x^2 + y^2)

Substituting these values into the angle bisector theorem equation, we have:

x/y = x/√(x^2 + y^2)

Cross-multiplying, we get:

x * √(x^2 + y^2) = x * y

Squaring both sides of the equation, we have:

x^2 * (x^2 + y^2) = x^2 * y^2

Expanding and simplifying, we get:

x^4 + x^2 * y^2 = x^2 * y^2

Subtracting x^2 * y^2 from both sides, we have:

x^4 = 0

Since x^4 = 0, we can conclude that x = 0.

Therefore, the length of segment AO is 0, which means that point A coincides with the origin O.

Now, let's consider angle A. Since point A coincides with the origin O, angle A is a zero angle, which is not less than 90°.

This contradicts our assumption that ΔABC is acute-angled. Therefore, our assumption is false, and ΔABC is not acute-angled.

Hence, we have proved that ΔABC is not acute-angled.

Note: The proof provided above is based on the given information and the angle bisector theorem.