Question

triangle abc and triangle on the same base ab.if line segment cd is bisect by ab at o . show that(abc)=ar(abd)triangle ABC and triangle A B D R two Triangles on the same base abf line segment CDS bisect by aviato show that area ABC equal to area AB de

Answer 1

Hello Mate!

Given : ∆ABC and ∆ABD are on same base and AB bisects CD.

To prove : ar(∆ABC) = ar(∆ABD)

Proof : In ∆CAD,

OC = OD

This means that OA is median and hence,

ar(∆AOC) = ar(∆AOD) __(1)

In ∆CBD,

Again, OC = OD

This means that OB is median and hence,

Hence, ar(∆BOC) = ar(∆BOD) __(2)

On adding (1) and (2) we get,

ar(∆AOC) + ar(∆BOC) = ar(∆AOD) + ar(∆BOD)

ar(∆ABC) = ar(∆ABD)

ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ.

Have great future ahead!

Answer 2

here is your answer OK ☺☺☺☺☺☺☺

I know this....,

ΔABC and Δ ADC are on the same base AB (and on the opposite sides of AB....)

In ΔDAC, AO is the median, (given, CD is bisected by AB)

ar(DAO) = ar(CAO) -----------{1} ( a median of a triangle divides it into 2 triangles of equal areas)

Similarlly, BO is the median of ΔCBD

∴ ar(DBO) = ar(CBO) -----------{2} ( a median of a triangle divides it into 2 triangles of equal areas)

Adding {1} and {2}

ar(DAO) + ar(DBO) = ar(CAO) + ar(CBO)

i.e., ar(ABD) = ar(ABC)

or, ar(ABC) = ar(ABD)

HENCE PROVED

Hope this Helps......