QUESTION 11
Easy
In ΔABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined (figure). Prove that
Area (ΔBPQ) = \(\frac{1}{2}\) Area (ΔABC).
SOLUTION
1
Write down what is given
D is the midpoint of AB.
P lies on BC.
Q lies on AB.
CQ || PD.
To prove: \(Area(ΔBPQ) = \frac{1}{2} Area(ΔABC)\)
P lies on BC.
Q lies on AB.
CQ || PD.
To prove: \(Area(ΔBPQ) = \frac{1}{2} Area(ΔABC)\)
2
Identify equal areas between parallel lines
Consider ΔDPQ and ΔDPC.
Both triangles share the same base PD and lie between same parallel lines PD and CQ. Therefore their areas are equal.
\(Area(ΔDPQ) = Area(ΔDPC)\) --- (1)
Both triangles share the same base PD and lie between same parallel lines PD and CQ. Therefore their areas are equal.
\(Area(ΔDPQ) = Area(ΔDPC)\) --- (1)
3
Express the area of ΔBPQ
From the given figure, the total area of ΔBPQ can be split into two smaller triangles:
\(Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ)\)
Substituting equation (1) in the above expression, we get
\(Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPC)\)
⇒ \(Area(ΔBPQ) = Area(ΔBDC)\)--- (2)
\(Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ)\)
Substituting equation (1) in the above expression, we get
\(Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPC)\)
⇒ \(Area(ΔBPQ) = Area(ΔBDC)\)--- (2)
4
Use the midpoint property
Since D is the midpoint of AB, the line segment CD is a median of ΔABC. A median always divides a triangle into two triangles of equal area. So,
\(Area(ΔBDC) = \frac{1}{2} Area(ΔABC)\) --- (3)
\(Area(ΔBDC) = \frac{1}{2} Area(ΔABC)\) --- (3)
5
Substitute equation (3) in (2) and prove
We get
\(Area(ΔBPQ) = \frac{1}{2} Area(ΔABC)\)
Hence proved.
\(Area(ΔBPQ) = \frac{1}{2} Area(ΔABC)\)
Hence proved.
Concept Note
1. If two triangles have the same base and lie between the same parallel lines, their areas are equal.
2. A median is a line segment connecting a vertex of a triangle to the midpoint of the opposite side. It divides a triangle into two smaller triangles of equal angles.