मराठी

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.

Advertisements
Advertisements

प्रश्न

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.

बेरीज
Advertisements

उत्तर

Given: In a parallelogram ABCD, P and Q are the mid-points of AS and CD, respectively.

To show: PRQS is a parallelogram.

Proof: Since, ABCD is a parallelogram.

AB || CD

⇒ AP || QC

Also, AB = DC


`1/2`AB = `1/2`DC   ...[Dividing both sides by 2]

⇒ AP = QC  ...[Since, P and Q are the mid-points of AB and DC]

Now, AP || QC and AP = QC

Thus, APCQ is a parallelogram.

∴ AQ || PC or SQ || PR  ...(i)

Again, AB || DC or BP || DQ

Also, AB = DC

⇒ `1/2`AB = `1/2`DC   ...[Dividing both sides by 2]

⇒ BP = QD  ...[Since, P and Q are the mid-points of AB and DC]

Now, BP || QD and BP = QD

So, BPDQ is a parallelogram.

∴ PD || BQ or PS || QR   ...(ii)

From equations (i) and (ii),

SQ || RP and PS || QR 

So, PRQS is a parallelogram.

Hence proved.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Quadrilaterals - Exercise 8.4 [पृष्ठ ८२]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
पाठ 8 Quadrilaterals
Exercise 8.4 | Q 7. | पृष्ठ ८२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.


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°.


ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH.


Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.


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.


In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: DC, if AB = 20 cm and PQ = 14 cm


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°


D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is ______.


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]


Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×