Question

State "Basic Proportionality Theorem" and use it to prove the following:

A line through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

Approach

State and use the Basic Proportionality Theorem (BPT) to prove that the line through the midpoint of one side of a triangle, parallel to another side, bisects the third side. Given: In triangle ABC, P is the midpoint of AB and PQ ∥ BC. To prove: Q is the midpoint of AC. By BPT, $\frac{A P}{PB} = \frac{A Q}{QC}$ . Since AP=PB, the ratio equals 1, so AQ=QC, hence Q is the midpoint.

Step-by-step working

1
Statement of BPT: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio
1 mark
2
Given: In $△ A BC$ , P is midpoint of AB and $PQ ∥ BC$ . To prove: Q is the midpoint of AC. Draw a correct labeled figure
1 mark
3
As $PQ ∥ BC$ , by BPT: $\frac{A P}{PB} = \frac{A Q}{QC}$
1 mark
4
Since $A P = PB$ (P is midpoint), $\frac{A P}{PB} = 1$
1 mark
5
$∴ \frac{A Q}{QC} = 1 \Rightarrow A Q = QC$ , so Q is the midpoint of AC
1 mark

Concepts used

basic proportionality theoremthales theoremprovemid-pointparallel linesbisects

Approach

Step-by-step working

Concepts used

Similar question types