In \(∆ABC\), the midpoint of \(BC\) is \(D\) (See Fig.). Median \(AD\) is drawn. \(P\) is any point on \(AD\). Show that \(area (∆ABP) = area (∆ACP)\).
Proof:
\(D\) is the midpoint of \(BC\).
So, \(BD\) \(DC\)
Now consider triangles \(△PBD\) and \(△PCD\).
Therefore, \(Area(△PBD)=\) ---(1)
Now consider triangles \(△ABD\) and \(△ACD\).
They have \(BD=DC\) and the same height from \(A\) to line \(BC\).
Therefore, \(Area(△ABD)=\) ---(2)
Subtracting equation (1) from equation (2), we get
\(Area(△ABD)−\)\(=Area(△ACD)−\)
\(Area(△ABP)=Area(△ACP)\)
Hence proved.
Answer variants:
\(Area(△PBD)\)
\(Area(△ACD)\)
\(Area(△PCD)\)
\(=\)
\(>\)