Advertisements
Advertisements
प्रश्न
In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.
Advertisements
उत्तर
Join B and D, suppose AC and BD out at O.
Then OC = `1/2` AC
Now,
CQ = `1/4` AC
⇒ CQ = `1/2` `[1/2 AC ]`
= `1/2` × OC
In Δ DCO, P and Q are midpoints of DC and OC respectively
∴ PQ || PO
Also in Δ COB, Q is the midpoint of OC and QR || OB
∴ R is the midpoint of BC
APPEARS IN
संबंधित प्रश्न
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.
In the given figure, seg PD is a median of ΔPQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that `"PM"/"PR" = 1/3`.
[Hint: DN || QM]
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 the figure, give below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that:
(i) AQ // BS
(ii) DS = 3 Rs.
The side AC of a triangle ABC is produced to point E so that CE = AC. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively.
Prove that:
- 3DF = EF
- 4CR = AB
In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find DE, if AB = 8 cm
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
The diagonals of a quadrilateral intersect each other at right angle. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.
In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: A is the mid-point of PQ.
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.
