Question

Pls check if this is a correct solution of the question:

Abc and abd are 2 triangles on same base AB. CD is bisected by AB at O, show that ar(ABC)= ar(ADC)

Answer 1

$There \:is \:mistake \:in \:the \: question \:itself$

$we \:have \:to \:Show \: that \\ar(\triangle ABC) = ar(\triangle ABD)$

$\underline { \blue { Solution:}}$

$From \:the \:figure , \\In \:\triangle AOC; \:\triangle BOD,\\OA = OB \: [given ] \\OC = OD \\\angle {AOC} = \angle {BOD} \\( Vertically \: opposite \: angles )$

$\therefore \:\triangle AOC \cong \:\triangle BOD \\\blue { ( SAS \: Congruence ) }$

$AC = BD \: ( CPCT ) \\\angle {OAC} = \angle {OBD} \: (CPCT)$

But these are alternate interior angles for the lines AC , BD .

$\therefore AC \parallel BD$

$As \:AC = BD \:and \:AC \parallel BD ; \\ADBC \:is \: parallelogram .$

$AB \:is \:a \:diagonal \: of \: ABCD .$

$\triangle ABC \cong \:\triangle ABD$

( Diagonals divides a parallelogram into two congruent triangles )

$\therefore ar( \triangle ABC) = ar( \triangle ABD)$

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Answer 2

Yes, it is correct.

Answer:

In triangle ABC, AO is the median (CD is bisected by AB at O)

So, ar(AOC)=ar(AOD)..........(i)

Also,

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

So, ar(BOC) = ar(BOD)..........(ii)

Adding (i) and (ii),

We get,

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

⇒ ar(ABC) = ar(ABD)

Hence showed.