Isosceles triangle. Detailed theory with examples (2020)

Properties of an isosceles triangle.
Signs of an isosceles triangle.
Isosceles triangle formulas:
- side length formulas;
- equal sides length formulas;
- formulas for height, median, bisector of an isosceles triangle.

An isosceles triangle is a triangle whose two sides are equal. These parties are called lateral and the third party is basis.

AB = BC - lateral sides

AC - base

Isosceles triangle properties

The properties of an isosceles triangle are expressed in terms of 5 theorems:

Theorem 1. In an isosceles triangle, the angles at the base are equal.

Proof of the theorem:

Consider an isosceles Δ ABC with the foundation AS .

The sides are equal AB = Sun ,

Therefore, the angles at the base ∠ BАC = ∠ BCA .

Theorem on the bisector, median, height, drawn to the base of an isosceles triangle

Theorem 2. In an isosceles triangle, the bisector drawn to the base is the median and the height.
Theorem 3. In an isosceles triangle, the median drawn to the base is the bisector and the height.
Theorem 4. In an isosceles triangle, the height drawn to the base is the bisector and median.

Proof of the theorem:

Dan Δ ABC .
From point V let's hold the height BD.
The triangle is divided into Δ ABD and Δ CBD. These triangles are equal because their hypotenuse and common leg are equal ().
Direct AS and BD are called perpendicular.
B Δ ABD and Δ BCD ∠ BAD = ∠ BСD (from Theorem 1).
AB = BC - the sides are equal.
Parties AD = CD, since point D divides the segment in half.
Hence Δ ABD = Δ BCD.
The bisector, height and median are one segment - BD

Output:

The height of an isosceles triangle, drawn to the base, is the median and bisector.
The median of an isosceles triangle, drawn to the base, is the height and bisector.
The bisector of an isosceles triangle, drawn to the base, is the median and the height.

Remember! When solving such problems, lower the height to the base of the isosceles triangle. To divide it into two equal right-angled triangles.

Theorem 5. If three sides of one triangle are equal to three sides of another triangle, then such triangles are equal.

Proof of the theorem:

Given two Δ ABC and Δ A 1 B 1 C 1. Sides AB = A 1 B 1; BC = B 1 C 1; AC = A 1 C 1.

Proof by contradiction.

Let the triangles not be equal (otherwise the triangles were equal in the first attribute).
Let Δ A 1 B 1 C 2 = Δ ABC, whose vertex C 2 lies in the same half-plane with vertex C 1 relative to the straight line A 1 B 1. By assumption, the vertices C 1 and C 2 do not coincide. Let D be the midpoint of the segment C 1 C 2. Δ A 1 C 1 C 2 and Δ B 1 C 1 C 2 are isosceles with a common base C 1 C 2. Therefore, their medians A 1 D and B 1 D are heights. Hence, the lines A 1 D and B 1 D are perpendicular to the line C 1 C 2. A 1 D and B 1 D have different points A 1 and B 1, therefore, do not coincide. But through point D of straight line C 1 C 2 only one straight line perpendicular to it can be drawn.
From here we came to a contradiction and proved the theorem.

Signs of an isosceles triangle

If two angles in a triangle are equal.
The sum of the angles of a triangle is 180 °.
If in a triangle, the bisector is the median or height.
If in a triangle, the median is the bisector or height.
If in a triangle, the height is the median or bisector.

Isosceles triangle formulas

b- side (base)
a- equal sides
a - angles at the base
b

Side length formulas(grounds - b):

b = 2a \ sin (\ beta / 2) = a \ sqrt (2-2 \ cos \ beta)
b = 2a \ cos \ alpha

Equal side length formulas - (a):

a = \ frac (b) (2 \ sin (\ beta / 2)) = \ frac (b) (\ sqrt (2-2 \ cos \ beta))
a = \ frac (b) (2 \ cos \ alpha)

L- height = bisector = median
b- side (base)
a- equal sides
a - angles at the base
b - the angle formed by equal sides

Formulas for height, bisector and median, through side and angle, ( L):

L = a sin a
L = \ frac (b) (2) * \ tg \ alpha
L = a \ sqrt ((1 + \ cos \ beta) / 2) = a \ cos (\ beta) / 2)

Formula of height, bisector and median, through the sides, ( L):

L = \ sqrt (a ^ (2) -b ^ (2) / 4)

b- side (base)
a- equal sides
h- height

The formula for the area of a triangle in terms of height h and base b, ( S):

S = \ frac (1) (2) * bh

The calculation of the height of the triangle depends on the figure itself (isosceles, equilateral, versatile, rectangular). In practical geometry, complex formulas, as a rule, do not occur. It is enough to know the general principle of calculations so that it can be universally applicable for all triangles. Today we will introduce you to the basic principles of calculating the height of a figure, calculation formulas based on the properties of the heights of triangles.

What is height?

Height has several distinctive properties

The point where all the heights meet is called the orthocenter. If the triangle is pointed, then the orthocenter is inside the figure, if one of the corners is obtuse, then the orthocenter is usually outside.
In a triangle where one angle is 90 °, the orthocenter and vertex are the same.
Depending on the type of triangle, there are several formulas for how to find the height of a triangle.

Traditional computing

If p is half the perimeter, then a, b, c are the designation of the sides of the required figure, h is the height, then the first and simplest formula will look like this: h = 2 / a √p (pa) (pb) (pc) ...
In school textbooks, you can often find problems in which the value of one of the sides of the triangle and the value of the angle between this side and the base are known. Then the formula for calculating the height will look like this: h = b ∙ sin γ + c ∙ sin β.
When given the area of the triangle - S, as well as the length of the base - a, the calculations will be as simple as possible. The height is found by the formula: h = 2S / a.
When the radius of a circle circumscribed around a figure is given, we first calculate the lengths of its two sides, and then proceed to calculate the given height of the triangle. To do this, we use the formula: h = b ∙ c / 2R, where b and c are two sides of the triangle that are not the base, and R is the radius.

How to find the height of an isosceles triangle?

All sides of this figure are equivalent, their lengths are equal, therefore the angles at the base will also be equal. From this it follows that the heights that we draw on the bases will also be equal, they are also medians and bisectors at the same time. In simple terms, the height in an isosceles triangle divides the base in two. The triangle with a right angle, which turned out after drawing the height, will be considered using the Pythagorean theorem. Let's designate the side as a and the base as b, then the height h = ½ √4 a2 - b2.

How to find the height of an equilateral triangle?

The formula for an equilateral triangle (figures where all sides are equal in size) can be found based on previous calculations. It is only necessary to measure the length of one of the sides of the triangle and designate it as a. Then the height is deduced by the formula: h = √3 / 2 a.

How do I find the height of a right triangle?

As you know, the angle in a right-angled triangle is 90 °. The height lowered by one leg is at the same time the second leg. On them, the heights of the triangle with a right angle will lie. To obtain data on the height, you need to slightly transform the existing Pythagorean formula, denoting the legs - a and b, and also measuring the length of the hypotenuse - c.

Find the length of the leg (the side to which the height will be perpendicular): a = √ (c2 - b2). The length of the second leg is found using exactly the same formula: b = √ (c2 - b2). After that, you can start calculating the height of a triangle with a right angle, having previously calculated the area of the figure - s. Height value h = 2s / a.

Calculations with a versatile triangle

When a versatile triangle has sharp corners, the height dropped to the base is visible. If the triangle is with an obtuse angle, then the height can be outside the figure, and you need to mentally continue it in order to get the connection point of the height and base of the triangle. The easiest way to measure the height is to calculate it through one of the sides and the magnitude of the angles. The formula looks like this: h = b sin y + c sin ß.

Isosceles is such triangle, in which the lengths of its two sides are equal to each other.

When solving problems on a topic "Isosceles triangle" it is necessary to use the following well-known properties:

1. Angles lying opposite equal sides are equal to each other.
2. Bisectors, medians and heights drawn from equal angles are equal to each other.
3. The bisector, median and height, drawn to the base of the isosceles triangle, coincide with each other.
4. The center of the inscribed circle and the center of the circumscribed circle lie at the height, and therefore on the median and the bisector drawn to the base.
5. Angles that are equal in an isosceles triangle are always sharp.

A triangle is isosceles if it has the following signs:

1. The two angles of the triangle are equal.
2. The height matches the median.
3. The bisector coincides with the median.
4. The height coincides with the bisector.
5. The two heights of the triangle are equal.
6. The two bisectors of a triangle are equal.
7. The two medians of the triangle are equal.

Let's consider several tasks on the topic "Isosceles triangle" and we will give a detailed solution to them.

Objective 1.

In an isosceles triangle, the height drawn to the base is 8, and the base refers to the side as 6: 5. Find the distance from the apex of the triangle is the point of intersection of its bisectors.

Solution.

Let an isosceles triangle ABC be given (fig. 1).

1) Since AC: BC = 6: 5, AC = 6x and BC = 5x. VN - the height drawn to the base of the AC of the triangle ABC.

Since point H is the middle of the AC (by the property of an isosceles triangle), then HC = 1/2 AC = 1/2 6x = 3x.

BC 2 = BH 2 + HC 2;

(5x) 2 = 8 2 + (3x) 2;

x = 2, then

AC = 6x = 6 2 = 12 and

BC = 5x = 5 2 = 10.

3) Since the intersection point of the bisectors of the triangle is the center of the inscribed circle, then
OH = r. The radius of a circle inscribed in a triangle ABC is found by the formula

4) S ABC = 1/2 * (AC * BH); S ABC = 1/2 * (12 * 8) = 48;

p = 1/2 (AB + BC + AC); p = 1/2 (10 + 10 + 12) = 16, then OH = r = 48/16 = 3.

Hence VO = VN - OH; VO = 8 - 3 = 5.

Answer: 5.

Objective 2.

The bisector AD is drawn in the isosceles triangle ABC. The areas of triangles ABD and ADC are equal to 10 and 12. Find the area of a square, enlarged three times, built at the height of this triangle, drawn to the base of the AC.

Solution.

Consider triangle ABC - isosceles, AD - bisector of angle A (fig. 2).

1) Let's write the areas of the triangles BAD and DAC:

S BAD = 1/2 AB AD sin α; S DAC = 1/2 AC AD sin α.

2) Find the area ratio:

S BAD / S DAC = (1/2 AB AD sin α) / (1/2 AC AD sin α) = AB / AC.

Since S BAD = 10, S DAC = 12, then 10/12 = AB / AC;

AB / AC = 5/6, then let AB = 5x and AC = 6x.

AH = 1/2 AC = 1/2 6x = 3x.

3) From triangle ABN - rectangular according to the Pythagorean theorem AB 2 = AN 2 + BN 2;

25x 2 = VN 2 + 9x 2;

4) S A ВС = 1/2 AS ВН; S A B C = 1/2 6x 4x = 12x 2.

Since S A BC = S BAD + S DAC = 10 + 12 = 22, then 22 = 12x 2;

x 2 = 11/6; VN 2 = 16x 2 = 16 11/6 = 1/3 8 11 = 88/3.

5) The area of the square is equal to BH 2 = 88/3; 3 88/3 = 88.

Answer: 88.

Objective 3.

In an isosceles triangle, the base is 4 and the side is 8. Find the square of the height dropped to the side.

Solution.

In triangle ABC - isosceles BC = 8, AC = 4 (fig. 3).

1) VN - the height drawn to the base of the AC of the triangle ABC.

Since point H is the middle of the AC (by the property of an isosceles triangle), then HC = 1/2 AC = 1/2 4 = 2.

2) From a triangle VNS - rectangular according to the Pythagorean theorem VS 2 = VN 2 + NS 2;

64 = BH 2 + 4;

3) S ABC = 1/2 (AC BH), as well as S ABC = 1/2 (AM BC), then equate the right-hand sides of the formulas, we get

1/2 AC BH = 1/2 AM BC;

AM = (AC · BH) / BC;

AM = (√60 4) / 8 = (2√15 4) / 8 = √15.

Answer: 15.

Task 4.

In an isosceles triangle, the base and the height dropped to it are equal to 16. Find the radius of the circle circumscribed about this triangle.

Solution.

In triangle ABC - isosceles base AC = 16, BH = 16 - height drawn to the base of AC (fig. 4).

1) AH = HC = 8 (by the property of an isosceles triangle).

2) From a triangle VNS - rectangular according to the Pythagorean theorem

BC 2 = BH 2 + HC 2;

BC 2 = 8 2 + 16 2 = (8 2) 2 + 8 2 = 8 2 4 + 8 2 = 8 2 5;

3) Consider a triangle ABC: by the theorem of sines, 2R = AB / sin C, where R is the radius of a circle circumscribed about a triangle ABC.

sin C = BH / BC (from the VNS triangle by definition of sine).

sin C = 16 / (8√5) = 2 / √5, then 2R = 8√5 / (2 / √5);

2R = (8√5 √5) / 2; R = 10.

Answer: 10.

Task 5.

The length of the height drawn to the base of the isosceles triangle is 36, and the radius of the inscribed circle is 10. Find the area of the triangle.

Solution.

Let the isosceles triangle ABC be given.

1) Since the center of the inscribed circle in the triangle is the point of intersection of its bisectors, then O ϵ VN and AO is the bisector of angle A, and the current OH = r = 10 (fig. 5).

2) VO = VN - OH; BO = 36 - 10 = 26.

3) Consider the triangle ABN. By the theorem on the bisector of the angle of a triangle

AB / AN = VO / OH;

AB / AH = 26/10 = 13/5, then let AB = 13x and AH = 5x.

By the Pythagorean theorem, AB 2 = AN 2 + BH 2;

(13x) 2 = 36 2 + (5x) 2;

169x 2 = 25x 2 + 36 2;

144x 2 = (12 3) 2;

144x 2 = 144 9;

x = 3, then AC = 2 AH = 10x = 10 3 = 30.

4) S ABC = 1/2 * (AC * BH); S ABC = 1/2 * (36 * 30) = 540;

Answer: 540.

Task 6.

In an isosceles triangle, the two sides are 5 and 20. Find the bisector of the angle at the base of the triangle.

Solution.

1) Suppose the sides of the triangle are 5 and the base is 20.

Then 5 + 5< 20, т.е. такого треугольника не существует. Значит, АВ = ВС = 20, АС = 5 (fig. 6).

2) Let LC = x, then BL = 20 - x. By the theorem on the bisector of the angle of a triangle

AB / AC = BL / LC;

20/5 = (20 - x) / x,

then 4x = 20 - x;

Thus, LC = 4; BL = 20 - 4 = 16.

3) We use the formula for the bisector of the angle of a triangle:

AL 2 = AB AC - BL LC,

then AL 2 = 20 · 5 - 4 · 16 = 36;

Answer: 6.

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Since the height of an isosceles triangle, lowered to the base, is both a bisector and a median, therefore, it divides the base and the angle at the apex into two equal parts, forming a right-angled triangle with sides a and b / 2. From the Pythagorean theorem in such a triangle, you can find the base itself, and then calculate all the other possible data. (fig. 88.2) h ^ 2 + (b / 2) ^ 2 = a ^ 2 b = √ (a ^ 2-h ^ 2) / 2

To calculate the perimeter of an isosceles triangle, add the base or the above radical through the height to the two lateral sides. P = 2a + b = 2a + √ (a ^ 2-h ^ 2) / 2

The area of an isosceles triangle through the height and base, by definition, is calculated as half of their product. Replacing the base with the expression corresponding to it, we get the area through the height and side of the isosceles triangle. S = hb / 2 = (h√ (a ^ 2-h ^ 2)) / 4

In an isosceles triangle, not only the sides are equal, but also the angles at the base, and since they always add up to 180 degrees, any of the angles can be found knowing the other. The first angle is calculated by the cosine theorem given for equal sides, and the second can be found through the difference from 180. (Fig. 88.1) cos⁡α = (b ^ 2 + c ^ 2-a ^ 2) / 2bc = (b ^ 2 + a ^ 2-a ^ 2) / 2ba = b ^ 2 / 2ba = b / 2a cos⁡β = (a ^ 2 + a ^ 2-b ^ 2) / (2a ^ 2) = (2a ^ 2 -b ^ 2) / (2a ^ 2) α = (180 ° -β) / 2 β = 180 ° -2α

The central median and bisector dropped to the base coincide with the height, and the lateral medians, heights and bisectors can be found using the following formulas for isosceles triangles. To compute them in terms of height and side, you need to replace the base with its equivalent expression. (fig. 88.3) m_a = √ (2a ^ 2 + 2b ^ 2-a ^ 2) / 2 = √ (a ^ 2 + 2b ^ 2) / 2

The height lowered to the lateral side through the height lowered to the base and lateral side of an isosceles triangle. (fig. 88.8) h_a = (b√ ((4a ^ 2-b ^ 2))) / 2a = (√ (a ^ 2-h ^ 2) √ ((4a ^ 2-a ^ 2 + h ^ 2 ))) / 2a = √ ((a ^ 2-h ^ 2) (3a ^ 2 + h ^ 2)) / 2

Lateral bisectors can also be expressed in terms of the lateral side and the central height of the triangle. (fig. 88.4) l_a = √ (ab (2a + b) (a + ba)) / (a + b) = √ (a (a ^ 2-h ^ 2) (2a + √ (a ^ 2-h ^ 2))) / (a + √ (a ^ 2-h ^ 2))

The middle line is drawn parallel to either side of the triangle, connecting the midpoints of the sides in relation to it. Thus, it always turns out to be equal to half of the side parallel to it. Instead of an unknown base, you can substitute the radical used in the formula to find the midline through the height and side of an isosceles triangle (Fig. 88.5) M_b = b / 2 = √ (a ^ 2-h ^ 2) / 2 M_a = a / 2

The radius of a circle inscribed in an isosceles triangle starts from the point at the intersection of the bisectors and goes perpendicularly to either side. To find it through the height and side of the triangle, you need to replace the base in the formula with a radical. (fig. 88.6) r = 1/2 √ (((a ^ 2-h ^ 2) (2a-√ (a ^ 2-h ^ 2))) / (2a + √ (a ^ 2-h ^ 2) ))

The radius of a circle circumscribed around an isosceles triangle is also derived from the general formula by substituting the radical through the height and side instead of the base. (fig. 88.7) R = a ^ 2 / √ (3a ^ 2-h ^ 2)