मराठी

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 - Mathematics

Advertisements
Advertisements

प्रश्न

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

बेरीज

उत्तर


In ΔBDF and ΔDRC,
BD = DC    ...(D is the mid-point of BC)
CR || PD || AB
∠BFD = DRC  ...(alternate angles)
∠BDF = RDC  ...(vertivally opposite angles)
Therefore,
ΔBDF ≅ ΔDRC
⇒ DF = DR    .....(i)

In ΔABC,
D is the mid-point of BC and DP || AB
Therefore, P is the mid-point of AC.

In ΔDEP,
C is the mid-point of PE and DP || RC || AB   ...(CE = `(1)/(2)"AC"` and P is the mid-point of AC)
Therefore, R is the mid-point of DE.
⇒ DR = RE ......(ii)
But EF = DF + DR + RE
EF = DF + DF + DF
EF = 3DF.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 15: Mid-point and Intercept Theorems - Exercise 15.1

APPEARS IN

फ्रँक Mathematics [English] Class 9 ICSE
पाठ 15 Mid-point and Intercept Theorems
Exercise 15.1 | Q 17.1
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×