Answer:
Theorem : Opposite sides of a parallelogram are congruent or equal.
Let us suppose a parallelogram ABCD.
Given: and (According to the definition of parallelogram)
We have to prove that: AB is congruent to CD and BC is congruent to AD.
Prove: let us take two triangles, and
In these two triangles, { By the definition of alternative interior angles}
Similarly,
And, AC=AC (common segment)
By ASA,
thus By the property of congruent triangle, we can say that corresponding sides of are also congruent.
Thus, AB is congruent to CD and BC is congruent to AD.
Step-by-step explanation:
Answer:
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Step-by-step explanation:
Given that:
In ∆AMN and ∆TPE
MA = PE (Side )
∠AMN = ∠TPE (angle)
MN = PT (Side)
By SAS Property
∆ MAN =~ ∆ PET
∆ MAN =~ ∆ PET are congruent triangles .
<u>Answer:-</u> ∆ MAN =~ ∆ PET
<em>Additional</em><em> comment</em><em>:</em>
SAS Property:-
In two triangles , The two sides and the included angle are equal to the corresponding two sides and the included angle of the second triangle then they are congruent and this property is called Side-Angle-Side (SAS) Property.
<u>a</u><u>lso </u><u>read</u><u> similar</u><u> questions</u><u>:</u> I don't understand proof on geometry. How do you prove something that is given?
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Geometry Proof: Given: ∠1 ≅ ∠2 ∠3 ≅ ∠4 Prove: M is the midpoint of JK
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Answer:
to get you answer you would have to add 1 1/2 and 2 1/4, then your equation would be 60 ÷ 3 3/4 , then whatever your total is, multiply that by 40 and that would be your answer
Answer:
No
Step-by-step explanation:
You cannot conclude that ΔGHI is congruent to ΔKJI, because although you can see/interpret that there all the angles are congruent with one another, like with vertical angles (∠GIH and ∠KIJ) and alternate interior angles (∠H and ∠J, ∠G and ∠K), we don't know the side lengths.
All the angles could be congruent, but the sides might be different. For example, ΔGHI might be a bigger triangle than ΔKJI, which could make them similar to one another, but not congruent.
For something to be congruent to another, everything must be exactly the same.