Show that the diagonals of a parallelogram divide it into four triangles of equal area.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that
ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]
In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(i) ar (BDE) = 1/4 ar (ABC)
(ii) ar (BDE) = 1/2 ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 1/8 ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]