A line segment (DE) joining the midpoints of two sides of a triangle is parallel
to the third sid

This is  called a mid segment theorem in which the al ine segment (DE) joining the midpoints of two sides of a triangle is parallel to the third side.

Step-by-step explanation:

Suppose we have a triangle ABC. Then the midpoints can be located as D, E and F. If we join D , E and F another triangle is formed.

From the figure we can see that

AE≅ CE

AD≅DB

BF≅CF  

BECAUSE  all the given points are the midpoints which divide the lines into two equal halves.

If we increase the line DE to a point L we find out that DL is parallel to BC i.e. it does not meet at any point with BC. ( the two lines do not meet)

(1)

If we join C with L we find out that the the line DE is half in length to the line BC.

AS

AE= CE  (midpoints dividing into equal line segements.)

LE= DE

Triangle CEL= Triangle DEF

so

DL= BC

But DE = 1/2 DL

therefore

DE= 1/2 BC (2)

Therefore from 1 and 2 we find that a line segment (DE) joining the midpoints of two sides of a triangle is parallel to the third side

dont think so

step-by-step explanation:

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it