Question

the line joining the vertices A and C of two triangles ABC and BCD intersects BD in point . Prove that area (AABD) area (ABCD) = AO CO. ​

Answer 1

Answer:

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)

Answer 2

Answer:

Step-by-step explanation:

Answer

Now ΔAOE and ΔFOD are similar.

As ∠EOA=∠FOD [Opposite angle], ∠AEO=∠DFO [right-angles] and ∠EAO=∠FDO [AE∥FD and AD ].

So  

AE/DF=AO/OD    (1).

Now  

1. area(DBC) /area(ABC) =1/2×BC×AE÷1/2×BC×FD=AE/FD=AO/FD[by (1)]

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