मराठी

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

Advertisements
Advertisements

प्रश्न

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.

बेरीज
Advertisements

उत्तर

In the adjoining figure, ABCD is a || gm in which E and F are the midpoints  of AB and CD respectively.

Since E and F are midpoints of AB and CD respectively

∴ AE = BE =`1/2` AB

And CF = DF =`1/2` CD

But, AB = CD

∴ `1/2` AB = `1/2` CD

⇒ BE = CF

Also, BE || CF                [∵ AB || CD]

∴ PHBE is a parallelogram

BE = PH         ....(i)

⇒ AEPG is parallelogram

∴ AE = GP          ....(ii)

But is the midpoint of AB

∴ AE = BE         ...(iii)

from (i), (ii) and (iii)

⇒ GP = PH

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Quadrilaterals - Exercise 13.4 [पृष्ठ ६५]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 9
पाठ 13 Quadrilaterals
Exercise 13.4 | Q 19 | पृष्ठ ६५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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 below Fig, ABCD and PQRC are rectangles and Q is the mid-point of Prove thaT

i) DP = PC (ii) PR = `1/2` AC


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


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.


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.


Side AC of a ABC is produced to point E so that CE = `(1)/(2)"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 meets AC at point P and EF at point R respectively. Prove that: 3DF = EF


Side AC of a ABC is produced to point E so that CE = `(1)/(2)"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 meets AC at point P and EF at point R respectively. Prove that: 4CR = AB.


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: F is the mid-point of BC.


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: The line drawn through G and parallel to FE and bisects DA.


E is the mid-point of a median AD of ∆ABC and BE is produced to meet AC at F. Show that AF = `1/3` AC.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×