QUESTION 9
Easy
In ΔABC, the midpoint of BC is D(figure). Median AD is drawn. P is any point on AD. Show that
area(ΔABP) = area(ΔACP).
SOLUTION
Since \(AD\) is a median,
\(BD=DC\)
Triangles \(ABD\) and \(ACD\) have
(i) equal bases BD and DC,
(ii) the same altitude from A.
Hence,
\(Area(ΔABD) = Area(ΔACD)\)
Point \(P\) lies on \(AD\).
Inside triangle \(ABD\),
\(Area(ΔABD) = Area(ΔABP) \,+\, Area(ΔBPD)\)
Similarly,
\(Area(ΔACD) = Area(ΔACP) \,+\, Area(ΔCPD)\)
Triangles \(BPD\) and \(CPD\) have equal bases \(BD\,=\,DC\), and the same height from \(P\).
Therefore,
\(Area(ΔBPD)\,=\,Area(ΔCPD)\)
Subtracting equal areas from equal areas,
\(Area(ΔABP)\,=\,Area(ΔACP)\)
Hence proved.
\(BD=DC\)
Triangles \(ABD\) and \(ACD\) have
(i) equal bases BD and DC,
(ii) the same altitude from A.
Hence,
\(Area(ΔABD) = Area(ΔACD)\)
Point \(P\) lies on \(AD\).
Inside triangle \(ABD\),
\(Area(ΔABD) = Area(ΔABP) \,+\, Area(ΔBPD)\)
Similarly,
\(Area(ΔACD) = Area(ΔACP) \,+\, Area(ΔCPD)\)
Triangles \(BPD\) and \(CPD\) have equal bases \(BD\,=\,DC\), and the same height from \(P\).
Therefore,
\(Area(ΔBPD)\,=\,Area(ΔCPD)\)
Subtracting equal areas from equal areas,
\(Area(ΔABP)\,=\,Area(ΔACP)\)
Hence proved.
Concept Note
1. A median divides a triangle into two triangles of equal area.
2. Triangles having the same base and lying between the same parallels have equal areas.