Ex 6.2, 9

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Ex 6.2, 9 In △ABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on 𝐴𝐷. Show that area (△𝐴𝐵𝑃)=area(△𝐴𝐶𝑃). We know that Median divides a triangle into two equal area In ∆ ABC Since AD is the median ∴ Area ∆ ABD = Area ∆ ACD In ∆ PBC Here, PD is the median of ∆ PBC ∴ Area ∆ PBD = Area ∆ PCD Doing (2) – (1) Area ∆ ABD – Area ∆ PBD = Area ∆ ACD – Area ∆ PCD Area ∆ ABP = Area ∆ ACP Hence proved