All categories

3 years ago

Given △ABC, AD and BE are medians, AD ∩BE=G. Prove: GD/AG = DE/AB

Mathematics

1 answer:

saw5 [17]3 years ago

3 0

Answer:

See explanantion

Step-by-step explanation:

Since AD and BE are medians, then points D and E are midpoints of the sides BC and AC, respectively. Thus, segment DE is triangle's midline. By the Triangle midline theorem, DE is parallel to the side AB. Thus,

$\angle ABE\cong \angle BED;$
$\angle BAD\cong \angle EDA$

as alternate interior angles.

Moreover, $\angle BGA\cong \angle EGD$ as vertical angles.

This gives us that triangles ABG and DEG are similar by AAA postulate.

Hence,

$\dfrac{AB}{ED}=\dfrac{AG}{GD}$

You might be interested in

For all values of x

Yuliya22 [10]

Answer:

A.) gf(x) = 3x^2 + 12x + 9

B.) g'(x) = 2

Step-by-step explanation:

A.) The two given functions are:

f(x) = (x + 2)^2 and g(x) = 3(x - 1)

Open the bracket of the two functions

f(x) = (x + 2)^2

f(x) = x^2 + 2x + 2x + 4

f(x) = x^2 + 4x + 4

and

g(x) = 3(x - 1)

g(x) = 3x - 3

To find gf(x), substitute f(x) for x in g(x)

gf(x) = 3( x^2 + 4x + 4 ) - 3

gf(x) = 3x^2 + 12x + 12 - 3

gf(x) = 3x^2 + 12x + 9

Where

a = 3, b = 12, c = 9

B.) To find g '(12), you must first find the inverse function of g(x) that is g'(x)

To find g'(x), let g(x) be equal to y. Then, interchange y and x for each other and make y the subject of formula

Y = 3x + 3

X = 3y + 3

Make y the subject of formula

3y = x - 3

Y = x/3 - 3/3

Y = x/3 - 1

Therefore, g'(x) = x/3 - 1

For g'(12), substitute 12 for x in g' (x)

g'(x) = 12/4 - 1

g'(x) = 3 - 1

g'(x) = 2.

5 0

2 years ago

Solve each equation and check.show all work.

harina [27]

Answer:

-b=11

Step-by-step explanation:

Nothing else is equal to a number except its self

4 0

3 years ago

(7 + 41) ÷ 2 – 15 =? Find the solution for this question.

koban [17]

precedence is:

1 Parentheses (simplify inside 'em)

2 Exponents

3 Multiplication and Division (from left to right)

4 Addition and Subtraction (from left to right)

(7+41) = 48

48 ÷ 2 = 24

24 -15 = 9

6 0

2 years ago

Read 2 more answers

HELP! Which relation is a function? Select all that apply.

Shtirlitz [24]

Answer:

A,B,D

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Is anybody else here to help me ??

Akimi4 [234]

Answer:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

$\cot(x)+\tan(x)$

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

Step-by-step explanation:

I'm going to use $x$ instead of $\theta$ because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

$\cot(x)+\cot(\frac{\pi}{2}-x)$

I'm going to use a cofunction identity for the 2nd term.

This is the identity: $\tan(x)=\cot(\frac{\pi}{2}-x)$ I'm going to use there.

$\cot(x)+\tan(x)$

I'm going to rewrite this in terms of $\sin(x)$ and $\cos(x)$ because I prefer to work in those terms. My objective here is to some how write this sum as a product.