In , is the midpoint of . is any point on , and is a point on such that . is joined (Fig. 6.34). Prove that

Understand the Question A median divides a triangle into two triangles of equal area. Since is the midpoint of , is a median of , so . Triangles on the same base and between the same parallels are equal in area. Triangles and share the base with , so their areas are equal. By adding to both, we can show that . Step 1 · Relate to Since is the midpoint of , . Triangles and share the same altitude from vertex to base . Step 2 · Equate Areas of Triangles on the Same Base Between Parallels Triangles and share the same base and lie between the same parallel lines . Therefore Step 3 · Express From the figure, point lies on segment : Substituting from : Step 4 · Relate to and Conclude From the figure, point lies on segment : Comparing and : Using : Answer > Common Mistakes Parallel Line Condition: For two triangles between the same parallels to have equal area, they must share the same base (or equal bases). Here, both and share the base . Median Area Property: Confusing median with angle bisector. A median divides a triangle into two triangles of equal area because their bases are equal and they share the same vertex altitude.

Measuring Space: Perimeter and Area | Exercise 6.2

Question 11

In $\Delta\text{ABC}$ , $D$ is the midpoint of $\text{AB}$ . $P$ is any point on $\text{BC}$ , and $Q$ is a point on $\text{AB}$ such that $\text{CQ} \parallel \text{PD}$ . $\text{PQ}$ is joined (Fig. 6.34). Prove that

$\text{Area}(\Delta\text{BPQ}) = \dfrac{1}{2} \text{Area}(\Delta\text{ABC})$

Solution

Understand the Question

A median divides a triangle into two triangles of equal area. Since $D$ is the midpoint of $\text{AB}$ , $\text{CD}$ is a median of $\Delta\text{ABC}$ , so $\text{Area}(\Delta\text{DBC}) = \dfrac{1}{2} \text{Area}(\Delta\text{ABC})$ .
Triangles on the same base and between the same parallels are equal in area. Triangles $\Delta\text{PDQ}$ and $\Delta\text{PDC}$ share the base $\text{PD}$ with $\text{CQ} \parallel \text{PD}$ , so their areas are equal.
By adding $\text{Area}(\Delta\text{BDP})$ to both, we can show that $\text{Area}(\Delta\text{BPQ}) = \text{Area}(\Delta\text{DBC}) = \dfrac{1}{2} \text{Area}(\Delta\text{ABC})$ .

Step 1 · Relate $\text{Area}(\Delta\text{DBC})$ to $\text{Area}(\Delta\text{ABC})$

Since $D$ is the midpoint of $\text{AB}$ , $\text{BD} = \dfrac{1}{2}\text{AB}$ . Triangles $\Delta\text{DBC}$ and $\Delta\text{ABC}$ share the same altitude from vertex $C$ to base $\text{AB}$ . Diagram 1

\begin{aligned} \text{Area}(\Delta\text{DBC}) &= \dfrac{\text{BD}}{\text{AB}} \times \text{Area}(\Delta\text{ABC}) \\[0.6em] &= \dfrac{1}{2} \text{Area}(\Delta\text{ABC}) \quad \dots (1) \end{aligned}

Step 2 · Equate Areas of Triangles on the Same Base Between Parallels

Triangles $\Delta\text{PDQ}$ and $\Delta\text{PDC}$ share the same base $\text{PD}$ and lie between the same parallel lines $\text{CQ} \parallel \text{PD}$ . Diagram 2

Therefore $\text{Area}(\Delta\text{PDQ}) = \text{Area}(\Delta\text{PDC}) \quad \dots (2)$

Step 3 · Express $\text{Area}(\Delta\text{BPQ})$

From the figure, point $D$ lies on segment $\text{BQ}$ : Diagram 3

$\text{Area}(\Delta\text{BPQ}) = \text{Area}(\Delta\text{BDP}) + \text{Area}(\Delta\text{DPQ})$

Substituting $\text{Area}(\Delta\text{PDQ}) = \text{Area}(\Delta\text{PDC})$ from $(2)$ : $\text{Area}(\Delta\text{BPQ}) = \text{Area}(\Delta\text{BDP}) + \text{Area}(\Delta\text{PDC}) \quad \dots (3)$

Step 4 · Relate $\text{Area}(\Delta\text{BPQ})$ to $\text{Area}(\Delta\text{DBC})$ and Conclude

From the figure, point $P$ lies on segment $\text{BC}$ : Diagram 4

$\text{Area}(\Delta\text{DBC}) = \text{Area}(\Delta\text{BDP}) + \text{Area}(\Delta\text{DPC}) \quad \dots (4)$

Comparing $(3)$ and $(4)$ : $\text{Area}(\Delta\text{BPQ}) = \text{Area}(\Delta\text{DBC})$

Using $(1)$ : $\text{Area}(\Delta\text{BPQ}) = \dfrac{1}{2} \text{Area}(\Delta\text{ABC})$

Answer

$\text{Area}(\Delta\text{BPQ}) = \dfrac{1}{2} \text{Area}(\Delta\text{ABC})$

Common Mistakes

Parallel Line Condition: For two triangles between the same parallels to have equal area, they must share the same base (or equal bases). Here, both $\Delta\text{PDQ}$ and $\Delta\text{PDC}$ share the base $\text{PD}$ .
Median Area Property: Confusing median with angle bisector. A median divides a triangle into two triangles of equal area because their bases are equal and they share the same vertex altitude.

Question 11

More questions in Exercise 6.2