In , the midpoint of is (Fig. 6.32). Median is drawn. is any point on . Show that .

Understand the Question A median of a triangle connects a vertex to the midpoint of the opposite side, dividing the triangle into two smaller triangles of equal area (since both share the same base length and perpendicular altitude). In , is the median, so . In , is the median, so . Subtracting the area of from and from gives . Step 1 · Compare Areas of Triangles with Vertex P In , is the midpoint of , so is the median. Since a median divides a triangle into two triangles of equal area Step 2 · Compare Areas of Triangles with Vertex A In , is the midpoint of , so is the median. Step 3 · Subtract Area Equations Subtracting equation from equation Answer > Hence proved. Common Mistakes Assuming Congruence: Equal area does not imply that the two triangles are congruent. Triangles and have equal areas because they have equal bases () and share the same altitude from vertex . Incorrect Subtraction Order: Always subtract the smaller triangle area () from the larger triangle area () consistently on both sides.

Measuring Space: Perimeter and Area | Exercise 6.2

Question 9

In $\Delta\text{ABC}$ , the midpoint of $\text{BC}$ is $\text{D}$ (Fig. 6.32). Median $\text{AD}$ is drawn. $\text{P}$ is any point on $\text{AD}$ . Show that $\text{area}(\Delta\text{ABP}) = \text{area}(\Delta\text{ACP})$ .

Solution

Understand the Question

A median of a triangle connects a vertex to the midpoint of the opposite side, dividing the triangle into two smaller triangles of equal area (since both share the same base length and perpendicular altitude).
In $\Delta\text{ABC}$ , $\text{AD}$ is the median, so $\text{area}(\Delta\text{ABD}) = \text{area}(\Delta\text{ACD})$ .
In $\Delta\text{PBC}$ , $\text{PD}$ is the median, so $\text{area}(\Delta\text{PBD}) = \text{area}(\Delta\text{PCD})$ .
Subtracting the area of $\Delta\text{PBD}$ from $\Delta\text{ABD}$ and $\Delta\text{PCD}$ from $\Delta\text{ACD}$ gives $\text{area}(\Delta\text{ABP}) = \text{area}(\Delta\text{ACP})$ .

Step 1 · Compare Areas of Triangles with Vertex P

In $\Delta\text{PBC}$ , $\text{D}$ is the midpoint of $\text{BC}$ , so $\text{PD}$ is the median. Diagram 1

Since a median divides a triangle into two triangles of equal area $\text{area}(\Delta\text{PBD}) = \text{area}(\Delta\text{PCD}) \quad \dots (1)$

Step 2 · Compare Areas of Triangles with Vertex A

In $\Delta\text{ABC}$ , $\text{D}$ is the midpoint of $\text{BC}$ , so $\text{AD}$ is the median. Diagram 2

$\text{area}(\Delta\text{ABD}) = \text{area}(\Delta\text{ACD}) \quad \dots (2)$

Step 3 · Subtract Area Equations

Subtracting equation $(1)$ from equation $(2)$ $\text{area}(\Delta\text{ABD}) - \text{area}(\Delta\text{PBD}) = \text{area}(\Delta\text{ACD}) - \text{area}(\Delta\text{PCD})$

$\text{area}(\Delta\text{ABP}) = \text{area}(\Delta\text{ACP})$

Answer

$\text{area}(\Delta\text{ABP}) = \text{area}(\Delta\text{ACP})$

Hence proved.

Common Mistakes

Assuming Congruence: Equal area does not imply that the two triangles are congruent. Triangles $\Delta\text{ABD}$ and $\Delta\text{ACD}$ have equal areas because they have equal bases ( $\text{BD} = \text{DC}$ ) and share the same altitude from vertex $\text{A}$ .
Incorrect Subtraction Order: Always subtract the smaller triangle area ( $\Delta\text{PBD}$ ) from the larger triangle area ( $\Delta\text{ABD}$ ) consistently on both sides.

Question 9

More questions in Exercise 6.2