[*][b]Median:[/b] In a triangle ABC, the median from A to BC (where M is the midpoint of BC) divides the triangle into two [b]equal-area[/b] triangles: [math]S_{ABM}=S_{ACM};[/math] [/*][*][b]Centroid (G):[/b] The intersection of the three medians is called the centroid or center of gravity. Key properties:[br][list][*]The centroid divides each median in the ratio AG:GM=2 (from vertex to base).[/*][*]The three larger triangles with vertex at G and sides along the vertices of the original triangle have [b]equal areas[/b]: [br][/*][/list][/*]
In triangle ABC, M is the midpoint of BC. Prove that [math]S_{ABM}=S_{ACM}[/math].
Consider the equal base and equal height.
In triangle ABC, M is the midpoint of BC. If [math]S_{ABM}[/math]=24, find[math]S_{ABC}[/math]
Use the area ratio created by the median.
In triangle ABC, let G be the centroid. [br]If [math]S_{BCG}=15[/math] , [math]S_{ABC}=[/math]..................
In triangle ABC, let G be the centroid. [br]If [math]S_{ABC}=60[/math], the area of the triangle BCG is: .........
Construct the medians BN and CP.[br]Find the ratio of the areas of triangles GNC and GAP.