So, in a parallelogram you have two sets of parallel lines. Take one set, and one of other lines and continue it beyond that set of two lines. Now, the single line is known as a tranversal, and by the same-side interior angles therom, the consecutive angles you are talking about must be congruent. The argument is the same on the opposite side. This is a lot easier with a diagram, ask your math teacher sometime.
Yes. All hexagons have interior angles totalling 720 degrees, whether they are regular or irregular. The proof of this is fairly easy.
The sum of three consecutive odd numbers must be divisible by 3. As 59 is not wholly divisible by 3 the question is invalid. PROOF : Let the numbers be n - 2, n and n + 2. Then the sum is 3n which is divisible by 3. If the question refers to three consecutive numbers then a similar proof shows that the sum of these three numbers is also divisible by 3. Again, the question would be invalid.
EVERY three consecutive numbers add to a multiple of 3: Proof: numbers are n, n + 1 and n + 2. The total is 3n + 3 or 3(n + 1) This means that for any three consecutive numbers, the total is 3 times the middle number.
I am working on the same exact proof right now and i am lost
There are 6 angles in a hexagon, and each angle measures 120 degrees. 120 x 6, or 120+120+120+120+120+120=720.
Without a visual or more information, I'm guessing that the picture is of angles 1 and 2 that are consecutive (share an angle side) and a separate picture of consecutive angles 3 and 4. With that said: 1) angle 2 congruent to angle 3................1) given 2) angle 1 is supplementary to angle 2....2) If angles are next to each other --> supps angle 3 is supplementary to angle 4 3) angle 1 congruent angle 4..............3) If supps to congruents angles ---> congruent
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supplementary
I can't offer a full proof, but I can suggest some possibilities that will lead you to your proof. In a parallelogram, you can easily demonstrate that the angles formed by a cord extending between parallel lines and the parallel lines themselves, and that are formed on opposite sides of the cord, are equal. This will work for both pairs of triangles in the parallelogram, and can be applied to all of the angles at the corners of the parallelogram. This will lead you to demonstrating that the pairs of triangles "pointing" to each other (not adjacent pairs) are similar, and in fact congruent. From there it is not difficult to establish that the connected sections of the two interior cords are equal.
corresponding angles
In accordance with WolframAlpha: "Heron's Theorem "The area (delta) of a triangle with side lengths a, b and c and semiperimeter s is given by (delta) = sqrt(s * (s-a) *(s-b) * (s-c)) ... Proof source: Heron of Alexandria. Proposition 1.8 in Metrica. Ca. 60 A.D." So Proposition 1.8 of Metrica is the well known Heron's Formula for finding the area of a triangle given its sides.
The parallelogram law states that when two concurrent forces F1 &F2 acting on a body are represented by two adjacent sides of a parallelogram the diagonal passing through their point of concurrency represents the resultant force R in magnitude and direction
Congruent shapes are identical in size and angles
First draw a parallelogram. I cannot draw one here so I will have to describe the picture and you should draw it. Let ABCD be a parallelogram. I put A on the bottom left, then B on the bottom right, C on the top right and D on the top left. Of course the arguments must apply to an arbitrary parallelogram, but just so you can follow the proof, that is my drawing. Now draw a segment from A to C. It is a diagonal. AB is parallel to CD and AD is parallel to BD because a parallelogram is a quadrilateral with both pair of opposite sides parallel. Now ABC and CDA both form triangles. Let angles 1 and 4 be the angles created by the diagonal and angle BCD of the parallelogram. Angle 1 is above and angle 4 is below. Similarly, let angles 3 and 2 be created by the intersection of the diagonal and angle DAB or the original parallelogram. Now angles 1 and 2 are congruent as are 3 and 4 because if two parallel lines are cut by a transversal, the alternate interior angles are congruent. Next using the reflexive property AC is congruent to itself. Now triangle ABC is congruent to triangle CDA by Angle Side Angle (SAS). This means that AB is congruent to CD and BC is congruent to AD by corresponding parts of congruent triangles are congruent (CPCTC). So we are done!
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Vertical angles
2 Column Proof: If a transversal intersects two parellel lines, then alternate interior angles are congruent.