Mathematics
P is the mid-point of diagonal AC of quadrilateral ABCD. Prove that the quadrilaterals ABPD and CBPD are equal in area.
Answer
Given: ABCD is a quadrilateral with diagonal AC and P is its mid-point of AC (AP = PC).
To Prove: Area of quadrilateral ABPD = Area of quadrilateral CBPD
Construction: Join PB and DP.
Proof: In Δ ABC, P is the midpoint of AC.
The line segment BP is median. A median divides a triangle into two triangles of equal area.
∴ Ar.(Δ ABP) = Ar.(Δ BCP) …………….(1)
Similarly, in Δ ADC, P is the midpoint of AC.
The line segment DP is median. A median divides a triangle into two triangles of equal area.
∴ Ar.(Δ ADP) = Ar.(Δ DCP) ……………..(2)
Adding equations (1) and (2),
⇒ Ar.(Δ ABP) + Ar.(Δ ADP) = Ar.(Δ BCP) + Ar.(Δ DCP)
⇒ Area of quadrilateral ABPD = Area of quadrilateral CBPD
Hence, the quadrilaterals ABPD and CBPD are equal in area.
Related Questions
Find the area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.
In trapezium ABCD, side AB is parallel to side DC. Diagonals AC and BD intersect at point P. Prove that triangles APD and BPC are equal in area.
In triangle ABC, D is mid-point of AB and P is any point on BC. If CQ parallel to PD meets AB at Q, prove that:
2 x area (△ BPQ) = area (△ ABC)
In △ ABC, D is a point on side AB and E is a point on AC. If DE is parallel to BC, and BE and CD intersect each other at point O; prove that :
(i) area (△ ACD) = area (△ ABE)
(ii) area (△ OBD) = area (△ OCE)