Question

In figure ABC and ADC are two triangles on the same base ab if line segment CD is bisected by ab at a show that area ABC is equal to area in ABD.
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Answer 1

SOLUTION:-

Given:-

• ∆ABC and ∆ABD are on the same base and and AB bisects CD.

Need to show:-

• area (ABC) is equal to area in (ABD).

Proof:-

In ∆ ACD,

=> CO = OD

Therefore,

AO is the median of ∆ ACD.

Therefore,

ar(AOD) = ar(AOC) ••••1 [since median of the triangle divide it into two equal parts]

Similarly,

In ∆BCD,

Since, CO = OD

Therefore,

BO is the median of ∆BCD.

Therefore,

ar(BOC) = ar(BOD) •••2

On adding,eq. 1 and 2

We get;

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

ar(ABC) = ar(ADB)

Hence proved✔️

Answer 2

Given

ΔABC and ΔABD are two triangles on the same base AB.

To show :

ar(ABC)=ar(ABD)

Proof :

Since the line segment CD is bisected by AB at O. OC=OD.

In ΔACD, We have OC=OD.

So, AO is the median of ΔACD

Also we know that median divides a triangle into two triangles of equal areas.

∴ar(ΔAOC)=ar(ΔAOD) _______ (1)

Similarly , In ΔBCD,

BO is the median. (CD bisected by AB at O)

∴ar(ΔBOC)=ar(ΔBOD) _______ (2)

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

ar(ΔAOC)+ar(ΔBOC)=ar(ΔAOD)+ar(ΔBOD)

∴ar(ΔABC)=ar(ΔABD)