Prove that the triangles on the same base and between the same parallel lines are equal in area.
1 Answer
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
Topics from Math
Algebraic Equation
0
Algebraic Fraction
2
Circle
16
Construction
3
Statistics
12
Triangle and Quadrilateral
17
Trigonometry
12
Compound Interest
6
Cylinder and Sphere
3
HCF and LCM
9
Indices
5
Prism and Pyramid
5
Radical and Surd
0
Plane Surface
5
Compound Depreciation
3
Population Growth
1
Set
4
Tax and Money Exchange
0
Related Questions