Question 9

A median divides a triangle into two triangles of equal area.

Step 1 — Areas of triangles with vertex P

D is the midpoint of BC. So, the length of BD equals the length of DC. Let's consider $\Delta$PBD and $\Delta$PCD. They have equal base lengths, BD and DC. They share the same vertex P. The perpendicular distance from P to BC is their common height. Triangles with equal bases and same height have equal areas.

$\text{area}(\Delta\text{PBD}) = \text{area}(\Delta\text{PCD}) \quad \text{(Equation 1)}$

Step 2 — Areas of triangles with vertex A

D is the midpoint of BC. So, the length of BD equals the length of DC. Let's consider $\Delta$ABD and $\Delta$ACD. They have equal base lengths, BD and DC. They share the same vertex A. The perpendicular distance from A to BC is their common height. Triangles with equal bases and same height have equal areas.

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

Step 3 — Subtracting areas

We have two equations from the previous steps. Let's subtract Equation 1 from Equation 2.

$\text{area}(\Delta\text{ABD}) - \text{area}(\Delta\text{PBD}) = \text{area}(\Delta\text{ACD}) - \text{area}(\Delta\text{PCD})$

From the diagram, we can see the resulting triangles. $\Delta$ABD minus $\Delta$PBD gives $\Delta$ABP. $\Delta$ACD minus $\Delta$PCD gives $\Delta$ACP.

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

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

Answer

Hence proved.