Advertisements
Advertisements
प्रश्न
In the given figure, AD and CE are medians and DF // CE.
Prove that: FB = `1/4` AB.
Advertisements
उत्तर
Given AD and CE are medians and DF || CE.
We know that from the midpoint theorem,
If two lines are parallel and the starting point of the segment is at the midpoint on one side, then the other point meets at the midpoint of the other side.
Consider triangle BEC. Given DF || CE and
D is the midpoint of BC.
So F must be the midpoint of BE.
So, FB = `1/2`BE but BE = `1/2`AB
Substitute value of BE in the first equation, we get
FB = `1/4`AB
Hence Proved.
APPEARS IN
संबंधित प्रश्न
In the given figure, `square`PQRS and `square`MNRL are rectangles. If point M is the midpoint of side PR then prove that,
- SL = LR
- LN = `1/2`SQ
D, E, and F are the mid-points of the sides AB, BC and CA of an isosceles ΔABC in which AB = BC.
Prove that ΔDEF is also isosceles.
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 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.
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.
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: ∠EFG = 90°
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if ______.
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.
P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.
