Instruction

1

Let the bisector of AE is drawn to the base BC of isosceles

**triangle**ABC. Triangle AEB will be rectangular, since the angle bisector AE will be its height. Side AB is the hypotenuse of this**triangle**, and BE and AE for his legs.By the Pythagorean theorem AB^2) = (YZ^2)+(AE^2). Then (BE^2) = sqrt (AB^2)-(AE^2)). Since AE and median**of the triangle**ABC, BE = BC/2. Therefore, (BE^2) = sqrt (AB^2)-((BC^2)/4)).If you specify the base angle ABC, then from rectangular**triangle**the angle bisector AE equal AE = AB/sin(ABC). Angle BAE = BAC/2, since AE is the bisector. Hence, AE = AB/cos(BAC/2).2

Suppose now that the height of BK is conducted to the side AC. This height is neither a median nor a bisector

**of the triangle**. To calculate its length there is a formula Stewart.The perimeter**of a triangle**is the sum of the lengths of all its sides is P = AB+BC+AC. And properiter equal to half the sum of the lengths of all its sides: P = (AB+BC+AC)/2 = (a+b+c)/2 where BC = a, AC = b, AB = c.Stuart the formula for the length of the bisector drawn to the side c (i.e., AB), will be of the form: l = sqrt(4abp(p-c))/(a+b).3

From the formula of Stuart can be seen that the angle bisector drawn to the side b (AC), will have the same length as b = c.

# Advice 2 : How to find the median

Under the median of the triangle refers to the segment which connects one of the vertices of the triangle with the midpoint of the opposite side. The definition implies that every triangle has three medians.

You will need

- Knowledge of the lengths of the sides of the triangle.

Instruction

1

To calculate the length of the median formula is applied (see Fig. 1), where:

mc is the length of the median;

a, b, c be the sidelengths of a triangle.

mc is the length of the median;

a, b, c be the sidelengths of a triangle.

Note

The medians of a triangle have the following properties:

1) any of the three medians divides the original triangle into two equal size triangle;

2) All medians of a triangle have a common intersection point. This point is called the center of the triangle;

3) the Medians of a triangle divide it into 6 equal triangles. Equal are called geometric shapes with equal areas.

1) any of the three medians divides the original triangle into two equal size triangle;

2) All medians of a triangle have a common intersection point. This point is called the center of the triangle;

3) the Medians of a triangle divide it into 6 equal triangles. Equal are called geometric shapes with equal areas.

Useful advice

If the triangle is an isosceles triangle, its medians are equal. In addition, in such a triangle the medians coincide with the bisectors and heights.

The angle bisector is the ray that emanates from any vertex of a triangle and forming it divides the angle in half.

Under the height of a triangle is meant the segment, which is drawn from the vertex of the triangle perpendicular to the opposite side.

The angle bisector is the ray that emanates from any vertex of a triangle and forming it divides the angle in half.

Under the height of a triangle is meant the segment, which is drawn from the vertex of the triangle perpendicular to the opposite side.

# Advice 3 : How to find the length of medians of the triangle

Median of a triangle is a segment drawn from any vertex to the opposite side, it divides it into parts of equal length. The maximum number of medians in a triangle is three, the number of vertices and sides.

Instruction

1

Task 1.

In an arbitrary triangle ABD held the median BE. Find its length if it is known that the parties, respectively, equal to AB = 10 cm, BD = 5 cm and AD = 8 cm.

In an arbitrary triangle ABD held the median BE. Find its length if it is known that the parties, respectively, equal to AB = 10 cm, BD = 5 cm and AD = 8 cm.

2

Solution.

Apply the formula median expression across all sides of the triangle. This is a simple task, since all the lengths of the sides are known:

BE = √((2*AB^2 + 2*BD^2 - AD^2)/4) = √((200 + 50 - 64)/4) = √(46,5) ≈ 6,8 (cm).

Apply the formula median expression across all sides of the triangle. This is a simple task, since all the lengths of the sides are known:

BE = √((2*AB^2 + 2*BD^2 - AD^2)/4) = √((200 + 50 - 64)/4) = √(46,5) ≈ 6,8 (cm).

3

Task 2.

In the isosceles triangle ABD, the sides AD and BD are equal. Held the median from vertex D to the side of the BA, it is BA with the angle equal to 90°. Find the length of the median DH, if you know that BA = 10 cm, and the angle DBA equal to 60°.

In the isosceles triangle ABD, the sides AD and BD are equal. Held the median from vertex D to the side of the BA, it is BA with the angle equal to 90°. Find the length of the median DH, if you know that BA = 10 cm, and the angle DBA equal to 60°.

4

Solution.

To find the median, identify one and equal sides of the triangle AD or BD. For this we consider one of the right triangles, suppose BDH. From the definition of the median, it follows that BH = BA/2 = 10/2 = 5.

Find the side BD in the formula of the trigonometric properties of a right triangle - BD = BH/sin(DBH) = 5/sin60° = 5/(√3/2) ≈ 5,8.

To find the median, identify one and equal sides of the triangle AD or BD. For this we consider one of the right triangles, suppose BDH. From the definition of the median, it follows that BH = BA/2 = 10/2 = 5.

Find the side BD in the formula of the trigonometric properties of a right triangle - BD = BH/sin(DBH) = 5/sin60° = 5/(√3/2) ≈ 5,8.

5

Now there are two options for finding the median: the formula used in the first task or the Pythagorean theorem for a right triangle BDH: DH^2 = BD^2 - BH^2.

DH^2 = (5,8)^2 - 25 ≈ 8,6 (cm).

DH^2 = (5,8)^2 - 25 ≈ 8,6 (cm).

6

Task 3.

In an arbitrary triangle BDA carried out the three medians. Find the length if it is known that the height of the DK equal to 4 cm and divides the base into segments of length BK = 3 and KA = 6.

In an arbitrary triangle BDA carried out the three medians. Find the length if it is known that the height of the DK equal to 4 cm and divides the base into segments of length BK = 3 and KA = 6.

7

Solution.

To find the median of the necessary lengths of all sides. The length of the BA can be found from the conditions: BA = BH + HA = 3 + 6 = 9.

Consider a right triangle BDK. By the Pythagorean theorem find the length of the hypotenuse BD:

BD^2 = BK^2 + DK^2; BD = √(9 + 16) = √25 = 5.

To find the median of the necessary lengths of all sides. The length of the BA can be found from the conditions: BA = BH + HA = 3 + 6 = 9.

Consider a right triangle BDK. By the Pythagorean theorem find the length of the hypotenuse BD:

BD^2 = BK^2 + DK^2; BD = √(9 + 16) = √25 = 5.

8

Similarly, find the hypotenuse of a right triangle KDA:

AD^2 = DK^2 + KA^2; AD = √(16 + 36) = √52 ≈ 7,2.

AD^2 = DK^2 + KA^2; AD = √(16 + 36) = √52 ≈ 7,2.

9

According to the formula expressions using side find the median:

BE^2 = (2*BD^2 + 2*BA^2 - AD^2)/4 = (50 + 162 - 51,8)/4 ≈ 40, hence BE ≈ 6,3 (cm).

DH^2 = (2*BD^2 + 2*AD^2 - BA^2)/4 = (50 + 103,7 - 81)/4 ≈ 18,2, hence DH ≈ 4,3 (cm).

AF^2 = (2*AD^2 + 2*BA^2 - BD^2)/4 = (103,7 + 162 - 25)/4 ≈ 60, hence AF ≈ 7,8 (cm).

BE^2 = (2*BD^2 + 2*BA^2 - AD^2)/4 = (50 + 162 - 51,8)/4 ≈ 40, hence BE ≈ 6,3 (cm).

DH^2 = (2*BD^2 + 2*AD^2 - BA^2)/4 = (50 + 103,7 - 81)/4 ≈ 18,2, hence DH ≈ 4,3 (cm).

AF^2 = (2*AD^2 + 2*BA^2 - BD^2)/4 = (103,7 + 162 - 25)/4 ≈ 60, hence AF ≈ 7,8 (cm).