Median AD divides ΔABC into equal areas: Area(ΔABD) = Area(ΔACD). Similarly, in ΔPBC, median PD gives Area(ΔPBD) = Area(ΔPCD). Subtracting these two equations yields Area(ΔABP) = Area(ΔACP).
In ΔABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (ΔABP) = area (ΔACP).
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Since D is the midpoint of BC, AD is a median of triangle ABC.
A median divides a triangle into two triangles of equal area:
Area(ΔABD) = Area(ΔACD) … (Equation 1).
In triangle PBC, PD is also a median because D is the midpoint of BC.
Therefore:
Area(ΔPBD) = Area(ΔPCD) … (Equation 2).
Subtracting Equation 2 from Equation 1:
Area(ΔABD) – Area(ΔPBD) = Area(ΔACD) – Area(ΔPCD)
Area(ΔABP) = Area(ΔACP).
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/