Theorem 5: One Pair of Opposite Sides Parallel and Equal
Trending Questions
In figure , AB ∥ DE, AB = DE, AC ∥ DF and AC = DF . Prove that BC ∥ EF and BC = EF.
In ΔABC and ΔDEF , AB=DE, AB || DE, BC =EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively ( see the given figure). Show that Quadrilateral BEFC is a parallelogram.
In figure , AB ∥ DE, AB = DE, AC ∥ DF and AC = DF . Prove that BC ∥ EF and BC = EF.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
In ΔABC and ΔDEF , AB=DE, AB || DE, BC =EF and BC ⃦EF. Vertices A, B and C are joined to vertices D, E and F respectively ( see the given figure). Show that Quadrilateral ACFD is a parallelogram.
Prove that if the diagonals of a parallelogram are perpendicular, then it is a rhombus. [3 MARKS]
In ΔABC and ΔDEF , AB=DE, AB || DE, BC =EF and BC ⃦EF. Vertices A, B and C are joined to vertices D, E and F respectively ( see the given figure). Show that Quadrilateral ACFD is a parallelogram.
In ΔABC and ΔDEF , AB=DE, AB || DE, BC =EF and BC ⃦EF. Vertices A, B and C are joined to vertices D, E and F respectively ( see the given figure). Show that AC = DF .
In ΔABC and ΔDEF , AB=DE, AB || DE, BC =EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively ( see the given figure). Show that Quadrilateral ABED is a parallelogram.
- True
- False
In figure, AB ∥ DE, AB = DE, AC ∥ DF and AC = DF. Then, BC ∥ EF and BC = ___ .
- EF
- BE
- CF
- AD
- is a rhombus
- is a parallelogram
- is a rectangle
- None of the above
In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B, and C are joined to vertices D, E and F respectively ( see the given figure).
AC||DF
AC=DF
The relation between AB and DF can't be determined
None of these
In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B, and C are joined to vertices D, E and F respectively ( see the given figure). Then, quadrilateral ACFD:
is a rectangle
is a rhombus
is a parallelogram
None of the above
The quadrilateral formed by joining the mid – points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if
(A) PQRS is a rectangle
(B) PQRS is a parallelogram
(C) diagonals of PQRS are perpendicular
(D) diagonals of PQRS are equal
$ \mathrm{ABCD}$ is a trapezium in which $ \mathrm{AB} \left|\right| \mathrm{DC}, \mathrm{BD}$ is a diagonal and $ \mathrm{E}$ is the mid-point of $ \mathrm{AD}$. A line is drawn through $ \mathrm{E} $parallel to $ \mathrm{AB}$ intersecting $ \mathrm{BC}$ at $ \mathrm{F}$ (see Fig.) . Show that $ \mathrm{F} $is the mid-point of $ \mathrm{BC}.$