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
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
In Fig. below, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are
respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.
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
In the Figure, `square`ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and PQ = `1/2 ("AB" + "DC")`.
In the given figure, M is mid-point of AB and DE, whereas N is mid-point of BC and DF.
Show that: EF = AC.
In parallelogram ABCD, E is the mid-point of AB and AP is parallel to EC which meets DC at point O and BC produced at P.
Prove that:
(i) BP = 2AD
(ii) O is the mid-point of AP.
In parallelogram ABCD, P is the mid-point of DC. Q is a point on AC such that CQ = `(1)/(4)"AC"`. PQ produced meets BC at R. Prove that
(i) R is the mid-point of BC, and
(ii) PR = `(1)/(2)"DB"`.
In ΔABC, P is the mid-point of BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R. Prove that: BC = 4QR
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.
The diagonals AC and BD of a quadrilateral ABCD intersect at right angles. Prove that the quadrilateral formed by joining the midpoints of quadrilateral ABCD is a rectangle.
