Question

The vertex A of triangle ABC is joined to a point D on the side BC.The mid point of AD isE.prove that ar(∆BEC)=1/2ar(∆ABC)

Answer 1

Hello Mate!

Given : E is mid point on AD.

To prove : ar(∆BEC) = ½ ar(∆ABC)

Proof : Since AE = DE ( E was mid point )

Hence, BE is median in ∆ABD.

Since median divides triangles into two equal areas therefore,

ar(∆ABE) = ar(∆BED) __(i)

Again, since AE = DE ( E was mid point )

Hence, CE is median in ∆ACD.

Since median divides triangles into two equal areas therefore,

ar(∆AEC) = ar(∆CED) __(ii)

On adding (i) and (ii) we get,

ar(∆ABE) + ar(∆AEC) = ar(∆BED) + ar(∆CED)

ar(quad ABEC) = ar(∆BEC)

ar(quad ABEC) + ar(∆BEC) = ar(∆ABC)

ar(∆BEC) + ar(∆BEC) = ar(∆ABC)

2ar(∆BEC) = ar(∆ABC)

ar(∆BEC) = ½ ar(∆ABC)

ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ

Have great future ahead!

Answer 2

here is your answer OK dude ☺☺☺☺☺☺

Since AD is a median,

Ar.(ABD)= Ar.(ACD)= 1/2 Ar.(ABC)

In triangle ABD

Since BE is a median,

Ar.(BED)= 1/2 Ar.(ABD)

or Ar.(BED)= 1/4 Ar.(ABC) [SInce Ar.(ABD)= 1/2 Ar.(ABC)]

SImilarly,

Ar.(CED)= 1/4 Ar.(ABC)

Ar.(BEC)= Ar.(BED)+ Ar.(CED)

or Ar.(BEC)= 1/4 Ar.(ABC)+ 1/4 Ar.(ABC)

or Ar.(BEC)= 1/2 Ar.(ABC)