Given,
Triangle ABC and DBC are standing on the same base BC and between the same parallel lines AD and BC.
To prove: Area of triangle ABC = Area of triangle DBC
Construction: Through point C, CE is drawn parallel to BA which meets AD at E.
Proof:
Statements | Reasons |
ABCE is a parallelogram. |
By construction. |
Triangle ABC = \(\frac12\) parallelogram ABCE. | Diagonal AC bisects parallelogram ABCE. |
Triangle DBC = \(\frac12\) parallelogram ABCE. | Both the triangle and parallelogram are standing on the same base and between the same parallels. |
Therefore, Triangle ABC = Triangle DBC | From Statements(2) and (3), Halves of the same parallelogram. |
Proved