Advertisements
Advertisements
Question
E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC. [Hint: Join AC]
Advertisements
Solution
Given: E is the mid-point of the side AD of the trapezium ABCD with AB || DC.
Also, EF || AB.

To prove: That F is the mid-point of BC.
Construction: Join AC which intersect EF at O.
Proof: In triangle ADC, E is the mid-point of AD and EF || DC. ...[Since, EF || AB and DC || AB. So, AB || EF || DC]
O is the mid-point of AC and OF || AB.
Now, OF bisect BC. ...[Converse of mid-point theorem]
Or F is the mid-point of BC.
Hence proved.
APPEARS IN
RELATED QUESTIONS
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.

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`
In a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F.
Prove that: AC = 3AF.
In triangle ABC, the medians BP and CQ are produced up to points M and N respectively such that BP = PM and CQ = QN. Prove that:
- M, A, and N are collinear.
- A is the mid-point of MN.
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: QAP is a straight line.
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.
If L and M are the mid-points of AB, and DC respectively of parallelogram ABCD. Prove that segment DL and BM trisect diagonal AC.
In a parallelogram ABCD, M is the mid-point AC. X and Y are the points on AB and DC respectively such that AX = CY. Prove that:
(i) Triangle AXM is congruent to triangle CYM, and
(ii) XMY is a straight line.
In AABC, D and E are two points on the side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet the side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet the side BC at points M and N respectively. Prove that BM = MN = NC.
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ______.
