# 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

• 3

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

• -1

AB PARALLEL TO DC

ANGLE DAC=ANGLE BCA

ANGLE BAC=ANGLE DCA

• 1 • 1
By using sss@congruency rule
• -1
You can see the answers all are correct
But refer ncert
• -1 • 1 • 2 • 1
SAS rule
• 1
Sorry
• -1 • 1
They arecorrect
• -1
Try it
• -1

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,
AC = AC [COMMON]
BC=DA [GIVEN]
SO Triangle ABC and ADC are congruent by SSS congruence rule
# HENCE PROVED..... THANK YOU...

• -1
opposite sides of a parallelogram are equal and th diagonal is common so by sss congruence both are congruent
• -1
b dhdh sjjsjs
• -2
Both are correct

• 1
They are absolutely correct!

• 1
you can prove it easily by the first theorem of this chapter
• 0
Hope this helps you! Cheers! • 0 • 0 in the //gm ,
AC=AC(common)
AB=DC(opposite sides )
by SSS congruency,
Hence proved.
• 0
We have the theorem • 1
U CAN DO IT BY SAS RULE
• 1 • 0
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.
• 0 • 0
SAS RULE
• 0
this can be proved by SSS
• 0
no idea
• 1
By SSS congruence criterion we can prove they are congruent. • 1
you can prove it by the first theorem of this chapter or by SSS or SAS congurant rule
• 1
try it ,its easy
• 1
Here's the solution • 0
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....
• 0
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
• 1 • 0
no idea
• 0 • 0
Ok tnkx
• 0
• 0
Mansi is correct
• 2
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
• 0
its very simple see the answer on the top.
• 0
fgvb jsfgv

• 0
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,
AC = AC [COMMON]
BC=DA [GIVEN]
SO Triangle ABC and ADC are congruent by SSS congruence rule
# HENCE PROVED..... THANK YOU...

• 0 • 0
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)
AC=AC(common)
Therefore triangles ABC and ACD are congruent
Hence proved

Hope it helped
• 0
8.1 theorm
• 0
We have the theorem • 1
ITS REALLY EASY QUESTION
ITS A TEXT BOOK QUESTION