Question

state the basic proportionality theorem.

Answer 1

If a line is drawn parallel to one side of a triangle intersectiong the other two sides , then it divides the two sides in the same ratio

Answer 2

Basic Proportionality Theorem..

Statement..
If a line parallel to a side of a triangle and intercept the remaining sides in two distinct points , then the line divides the side in the same proportion.

Given..
In triangle XYZ, line AB || YZ
Line AB intersects side XY and side XZ in points A and B respectively such that X-A-Y
and X-B-Z

To prove..
$\frac{xa}{ay} = \frac{xb}{bz}$
Construction..
Draw seg BY and AZ.

Proof..
Triangle XAB and Triangle BAY have common vertex B and their bases XA and AY line on the same line XY.
So,They have equal heights.
$\frac{a(triangle \: xab)}{a(triangle \: bay)} = \frac{xa}{ay}$
Now,
There
Triangle xab and Triangle ABZ have a common vertex A and their bases XB and BZ lie in the same line.
$\frac{a(triangle \: xab)}{a(triangle \: abz)} = \frac{xb}{bz}$
Triangle BAY and Triangle ABZ lie between the same two parallel lines AB and YZ.
So , they have equal heights,also they have same base AB.
A (triangle BAY)= A ( triangle ABZ)
Therefore...
$\frac{a(triangle \: xab)}{a(triangle \: bay) } = \frac{a(triangle \: xab)}{a(triangle \: abz)}$
Hence..we get
$\frac{xa}{ay} = \frac{xb \:}{bz}$
This is the answer
Hope it helps you.