Problem
GEO-B1-M06-P016 A Point on a Diagonal of a Parallelogram
#16
★★★☆☆ Level 3 of 5
In parallelogram \(ABCD\), point \(P\) lies on diagonal \(AC\). Prove that \(S_{ABP}=S_{ADP}\).
Compare the fractions of areas of triangles \(ABC\) and \(ACD\) cut by point \(P\) on the common diagonal.
Let \(AP:PC=t:(1-t)\). In triangle \(ABC\), \(S_{ABP}:S_{PBC}=AP:PC\), so \(S_{ABP}=tS_{ABC}\). Similarly, in triangle \(ACD\), \(S_{ADP}=tS_{ACD}\). But a diagonal of a parallelogram divides it into two equal-area triangles: \(S_{ABC}=S_{ACD}\). Hence \(S_{ABP}=S_{ADP}\).
A good example of area chasing on a diagonal of a parallelogram.