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Prove that: A Diagonal of a parallelogram divides it into two congruent triangles.

 its vry simple

see 

there can b two pairs of alternae interior angles equal as it is a parallegram.

and one side is common

there fore the two triangles are congruent by ASA criteria.

therefore a diagonal of a parallelogram divides it into two congruent triangles

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ITS REALY EASY

OPPOSITE SIDES OF A PARALLELOGRAM WILL BE EQUAL AND ONE PAIR OF ALTERNATE INTERIOR ANGLE WILL BE EQUAL SO TE TRIANGLES ARE CONGRUENT BY SAS CRITERIA

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AB PARALLEL TO DC

AD PARALLEL TO BC

ANGLE DAC=ANGLE BCA

ANGLE BAC=ANGLE DCA

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Here the answer

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By using sss@congruency rule
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You can see the answers all are correct
But refer ncert
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Please find this answer

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Please find this answer

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Please find this answer

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SAS rule
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Sorry
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Please find this answer

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They arecorrect
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Try it
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in IIgm opp. sides are equal so in a llgm ABCD , AB = CD and BC= DA.
So , now draw a diagonal AC inside the llgm  then, 
In triangle ABC and ADC
AB=AD [GIVEN]
AC = AC [COMMON]
BC=DA [GIVEN]
SO Triangle ABC and ADC are congruent by SSS congruence rule 
# HENCE PROVED..... THANK YOU...

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opposite sides of a parallelogram are equal and th diagonal is common so by sss congruence both are congruent
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b dhdh sjjsjs
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Both are correct
 
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They are absolutely correct!
 
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you can prove it easily by the first theorem of this chapter
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Hope this helps you! Cheers!

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Please find this answer

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 in the //gm ,
AC=AC(common)
AB=DC(opposite sides )
AD=BC(opposite sides )
by SSS congruency,
Hence proved.
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We have the theorem

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U CAN DO IT BY SAS RULE
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Please find this answer

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given:AC is diagonal of parallelogram ABCD
to prove: triangle ABC and ACD are congurent
PROOF: In triangle ABC and CDA
              angle 1 = angle 3 [alternate angles]
              angle 2 = angle 4 [alternate angles]
              AC=AC                 [common]
              by ASA rule
              triangle ABC is congurent to triangle CDA.
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I hope this is helpful.

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SAS RULE
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this can be proved by SSS
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no idea
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By SSS congruence criterion we can prove they are congruent.

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you can prove it by the first theorem of this chapter or by SSS or SAS congurant rule
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try it ,its easy
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Here's the solution

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Take IIgm ABCD and diag. AC.
AC=AC
AB=CD
BC=DA
Therefore, TRIANGLE ABC is congruent to TRIANGLE CDA.... By SSS cong. Criteria
HENCE PROVED>
Same can be done with diag. BD....
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Given: A parallelogram ABCD and AC is its diagonal .
To prove :?ABC??CDA
Proof :In??ABC and ?CDA ,
AC. =. AC
:- BY ASA congrency rule ,
?ABC. =. ?CDA
OK
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this is your proof

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no idea
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Completed answer

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Ok tnkx
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Buy an rd sharma search for the answer
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Mansi is correct
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abcd is a parallelogram and ac is the diagonal .
so in triangle abc and triangle cda
ab = cd      (as they are opposite sides of a parallelogram)
bc = da       (opposite sides of a parallelogram)
ac = ac       (common)
   therefore    triangle abc  is congruent to  triangle cda
area of congruent triangles are equal 
so area of abc is equal to area of cda
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its very simple see the answer on the top.
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fgvb jsfgv
 
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in IIgm opp. sides are equal so in a llgm ABCD , AB = CD and BC= DA.
So , now draw a diagonal AC inside the llgm  then, 
In triangle ABC and ADC
AB=AD [GIVEN]
AC = AC [COMMON]
BC=DA [GIVEN]
SO Triangle ABC and ADC are congruent by SSS congruence rule 
# HENCE PROVED..... THANK YOU...
 
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Please find this answer

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Proving theorom
Given:ABCD is a parallelogram.
Diagonal AC is drawn.
RTP:Triangle ABC and triangle ACD are congruent
Proof:Consider triangles ABC and ACD
AB=CD(opposite sides of parallelogram)
AD=BC(as given above)
AC=AC(common)
Therefore triangles ABC and ACD are congruent
Hence proved

Hope it helped
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8.1 theorm
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We have the theorem

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ITS REALLY EASY QUESTION  
ITS A TEXT BOOK QUESTION 
PLEASE REFER TO NCERT
OR THE ANSWERS GIVEN ABOVE IS ALSO CORRECT ANSWERS 




THANKS 
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