|A straight line drawn parallel to a side of a triangle, divides the other two sides proportionally.
Data: Triangle ABC, D is a point on AB and the line DE is parallel to BC
To prove: AD/BD = BE/CE
Consider triangle ABC and triangle BDE
|Step||Statement|| Reason |
|1||ang BAC = ang BDE||Corresponding angles because DE ll AC|
|2||ang BCA = ang BED||Corresponding angles because DE ll AC|
|3||ang DBE = ang ABC||Common angle|
|4||triangle ABC lll triangle BDE||AAA postulate on similarity|
|5||AB/BD = BC/BE||corresponding sides are proportional|
|6||(BD+DA)/BD =(BE+EC)/BE||Split AB and AC|
|7||1+AD/BD = 1+CE/BE||Simplification|
|8||AD/BD = CE/BE||Subtract 1 from either sides|
We shall try to prove the converse of basic proportionality theorem.
Let us consider triangle ABC shown in the figure below.
Divide the sides of the triangle AB and AC into any number of equal parts.
By measuring the angles B1 and B we observe that they are equal.
But B1 and B are corresponding angles.
Hence B1C1 ll BC.
AB2/B2 B= AC2/C2C = 2/3 and
B22C2 ll BC
AB3/B3B =AC3/C3C= 3/2 and B3C3 ll BC
AB4/B4B = AC4/C4C and B4C4 ll BC
Example: In the figure, find the length of PS given that ST ll QR.
Let PS = x cm
As ST ll QR
PS/QS = RT/PT
x/3 = 3/2
x = 9/2
Answer: PS = 4.5 cms
Directions: Answer the following questions.