MATH SOLVE

2 months ago

Q:
# In the given figure, ABCD is a parallelogram. E is any point on AB. CB and DE are produced to meet at F. Prove that area of triangle CEF = area of triangle BDF. if AB=EB, then area triangle BCE = area of triangle BDE = 1/4 area of parallelogram ABCD.

Accepted Solution

A:

Answer:Keep these axioms in mind while proving this
1. If a triangle and parallelogram have same base and are between the same parallel, then area of Parallelogram= × (Area of triangle)
2. Two parallelogram sharing the same base and if the sides parallel to the base are lying in the same line, their areas are equal.
Given: A B CD is a parallelogram and E is any point on AB . If DE produced meets CB produced at F .
Proof:
AS, A B CD is a parallelogram.
AB ║CD
AD ║CB
C F ║AD
Δ A D F and ║gm ABCD have same base AD and lying between the same parallels, AD ║CF.
→→Area ( Δ A D F)= × Area (║gm ABCD)------(1)
Similarly, Δ D E C and ║gm ABCD have same base CD and lying between the same parallels, CD ║AB.
→→Area ( Δ D E C)= × Area (║gm ABCD)-----(2)
From (1) and (2)
→→→→Area ( Δ A D F)=Area ( Δ DEC)
(2) Area ( Δ A D F)= Area ( Δ DEC)
→→Area ( Δ A D F)- Area (ΔA DE)= Area ( Δ DEC)- Area (ΔA DE)
→→Area ( Δ A E F)=Area ( Δ A D E)+Area ( Δ B E C)-Area (ΔA DE)→→[Area ( Δ D E C)= × Area (║gm ABCD)]→→→[ Area ( Δ D E C)=Area ( Δ A D E)+Area ( Δ B E C)]
→→Area ( Δ A E F)=Area ( Δ B E C)
Step-by-step explanation: