Advertisements
Advertisements
Question
ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: 2EF = BD.
Advertisements
Solution
In ΔBCD, E and F are the mid-points of DC and BC respectively.
Also EF || BD
Therefore, EF = `(1)/(2)"BD"`
⇒ 2EF = BD.
APPEARS IN
RELATED QUESTIONS
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
- D is the mid-point of AC
- MD ⊥ AC
- CM = MA = `1/2AB`
ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively
intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of
ΔABC
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.
D, E, and F are the mid-points of the sides AB, BC, and CA respectively of ΔABC. AE meets DF at O. P and Q are the mid-points of OB and OC respectively. Prove that DPQF is a parallelogram.
In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.
In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
In ΔABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that: A is the mid-point of PQ.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: ΔGEA ≅ ΔGFD
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: EGFH is a parallelogram.
In ΔABC, D and E are the midpoints of the sides AB and AC respectively. F is any point on the side BC. If DE intersects AF at P show that DP = PE.
