Q. 25.0( 2 Votes )

D, E and F are the mid-points of, and respectively in ΔABC. Prove that the area of BDEF = 1/2 area of ΔABC.

Answer :

We have

Given: D is the midpoint of , E is the midpoint of & F is the midpoint of .

AF = FB, AE = EC & BD = DC

To Prove: Area of BDEF = 1/2 Area of ∆ABC

Proof: Since, given is that D, E and F are midpoints of respectively.

We know, AE = EC.

AE = FD [, EC = FD]

[, ] …(i)

Also, we know that, AF = FB.

AF = ED [, FB = ED]

[, ] …(ii)

Now, in ∆AFE and ∆DEF, we have


By SSS theorem of congruence, we can say that the congruency AFE DEF is a similarity.

Similarly, in ∆AFE and ∆FBD, we have


By SSS theorem of congruence, we can say that the congruency AFE FBD is a similarity.

Similarly, the correspondence AEF EDC is a similarity.

So, ∆AFE, ∆DEF, ∆FBD and ∆EDC are all congruent triangles.

∆AFE = ∆DEF = ∆FBD = ∆EDC …(iii)



BDEF = ∆AFE + ∆AFE [from equation (iii)]

BDEF = 2 ∆AFE …(iv)


∆ABC = ∆AFE + ∆DEF + ∆FBD + ∆EDC

∆ABC = ∆AFE + ∆AFE + ∆AFE + ∆AFE [from equation (iii)]

∆ABC = 4 ∆AFE

∆ABC = 2 (2 ∆AFE) …(v)

Comparing equations (iv) and (v), we get

∆ABC = 2 (BDEF)

BDEF = 1/2 ∆ABC

Area of BDEF = 1/2 (Area of ∆ABC)

Thus, proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Basic Proportionality Theorem42 mins
Quiz | Criterion of Similarity of Triangle45 mins
Champ Quiz | Thales Theorem49 mins
NCERT | Strong Your Basics of Triangles39 mins
RD Sharma | Imp. Qs From Triangles41 mins
R.D Sharma | Solve Exercise -4.2 and 4.3FREE Class
How to Ace Maths in NTSE 2020?36 mins
R.D Sharma | Solve Exercise-4.545 mins
NCERT | Basic Proportionality Theorem22 mins
RD Sharma | Imp Qs Discussion- Triangles43 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses