Saved Bookmarks
| 1. |
52. In the given figure, ABCD is a kite in whichAB AD and O is the mid-point of EG and FH. Thequadrilateral EFGH is always a(1) Parallelogram(3) Square(2) Rectangle(4) Rhombus |
|
Answer» 1 Secondary School Math 8 points ABCD is a kite having AB = AD and BC= CD. prove that the figure formed by joining the mid - points of the sides , in order , is a rectangle. Advertisement Ask for details Follow Report byKenypiyupriyaagarw30.10.2016 Answers  prabhjyot Ambitious ABCD is a kite and AB=AD and BC=CD.EFGH are mid points of AB ,BC,CD,AD resp.To prove that EFGH is a rectangle. Construction: join AC and BD.In ΔABD , E and F are mid points.so we can write ,EF ║BD andEF = 1/2 BD ---------(1) (using mid point theorem) Now , in Δ BCD , G and H are mid points.So, GH║ BD andGH=1/2 BD ------------>(2) (by mid point th.)From (1) and (2) EF║ GH and EF = GH ( opposite sides of quad. EFGH are parallel and equal so EFGH is a parallelogram) Now since ABCD is a kite ,diagonal intersect at 90.So ∠AOd= 90. |
|