Questions on Geometry: Proofs in Geometry answered by real tutors!

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since AC=CE
and the 2 angles are the same then DC=AC making triangle DCE an isosceles triangle
so if BC=CE then AB=ED
thats the best i can figure this out with what information is given

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write the proof of the following theorem: if a quadrilateral is a kite,then its diagonals are perpendicular:

 
Given:  Kite ABCD
To prove: AC ⊥ BD
 
  

A couple of reasons are filled in.  You fill in the rest:


 1. AB ≅ AD
 2. BC ≅ CD
 3. AC ≅ AC 
 4. ᐃABC ≅ ᐃADC            4. SSS  (using 1,2, and 3)
 5. ∠BAE ≅ ∠DAE
 6. ᐃABD is isosceles
 7. ∠ABE ≅ ∠ADE
 8. ᐃABE ≅ ᐃADE            5. ASA  (Using 5,1, and 7)  
 9. ∠AEB ≅ ∠AED
10. ∠AEB ≅ ∠AED
11. m∠AEB = m∠ADE = 90°
12. AE ⊥ BD
13. AC ⊥ BD   
 
Edwin

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Assume the opposite of what you are trying to prove is true, find a contradiction.

For example, if you wanted to prove that there are infinitely many prime numbers, first assume that there are a finite number of primes. If you let N = 2*3*5*7*...*p where p is the "largest" prime, then N+1 must either be prime or be the product of two primes. However, N+1 cannot be divisible by any prime less than or equal to p, so if N+1 is composite, a larger prime divides it --> contradiction.

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Did you post the whole question? Because it's sorta, kinda, impossible to prove Z = X+Y without any other information.

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Assume that, on the contrary, triangle RST has a right angle. Then it must have a right angle and an obtuse angle. The sum of these two angles would have to be greater than 180 (90 plus a number greater than 90), which implies a contradiction. Hence RST cannot have any right angles.

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help me figure out proofs involving triangles?
Given: m∠3 = m∠7
m∠2 = 60°
m∠6 = 115°

Since m∠2 = 60°, m∠1 = 180°-60° = 120°

Since m∠6 = 115°, m∠5 = 180°-115° = 65°

That's all you wanted, but I'm going to find them all, since
no doubt you'll need to know how.

Since m∠5 = 65° and m∠2 = 60°, and angles 2,3, and 5 are interior angles
of the same triangle, and have sum 180°, then m∠3 = 180°-65°-60° = 55°

m∠7 = 55°, since we are given that m∠3 = m∠7

Since m∠6 = 115° and m∠7 = 55°, and angles 4,6, and 7 are interior angles
of the same triangle, and have sum 180°, then m∠4 = 180°-115°-55° = 10°
(You got that one right, but you didn't need it to find m∠5)

m∠1 = 120°
m∠2 = 60°
m∠3 = 55°
m∠4 = 10°
m∠5 = 65°
m∠6 = 115°
m∠7 = 55°

Edwin

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Was this an open-ended question?
There is more than one way to solve the problem.
The coordinates of the vertices of the triangle are irrelevant to proving ADEF is a parallelogram. Without giving coordinates, they could have told you that triangle ABC is not isosceles, and let you prove ADEF is not a rhombus using pure geometry.
The fact that they give you coordinates at all means that either they want to give alternate ways to solve it for those not as tuned into geometry, or (scary thought) that they expect you to solve the problem through analytical geometry, using a lot of arithmetic.
You could prove ADEF is a parallelogram through pure geometry first, and then you can use the coordinates to prove that ADEF is not a rhombus without much calculation. In fact, you could prove, for a general case, that a parallelogram so constructed is a rhombus if and only if the vertex conserved is the vertex of an isosceles triangle (in other words, if AC=AB).
It all seems obvious and simple in hindsight, and would be easy to explain, except that the "show your work" requirement has to be met following the conventions and formats favored by your teacher, and/or class, and/or textbook.
On the other hand, the arithmetic involved in the analytical geometry approach may be more tolerable than a pure geometry proof in a format that will satisfy the teachers or test graders.
I'll happily leave the choice to you, but I will give you two drawings to refer to.
For pure geometry--> and <--for analytical geometry
THE GEOMETRY WAY
On connecting the midpoints of the sides of any triangle, you are constructing lines parallel to the sides.
so since the ratios are the same, lines AC and DE are parallel.
You can prove that the other pairs of lines are parallel in a similar manner.
That way you find out that DE is parallel to AC (or AD), and that FE is parallel to AB (or AF). So ADEF is a parallelogram.
You could also say that you are splitting the original triangle into 4 similar triangles.
AD and AF would be the same length and ADEF would be a rhombus, if and only if AB and AC were the same length, which could happen only if ABC were isosceles.
However, AD and AF are halves of the original triangle sides AC and AD, which are not the same length. So the lengths of AD and AF are not the same, and ADEF is not a rhombus. You prove that AB is longer than AC from the coordinates. The length of horizontal segment AC is the difference of the x coordinates.
AB=
The length of AB could be calculated invoking Pythagoras or the distance formula from the differences of x and y-coordinates for A and B, as
AB=
AB=
I can easily see that
AB==AC,
and so AB>AC, making AD>AF, and ADEF is not a rhombus.
ONE WAY TO USE ANALYTICAL GEOMETRY
Another way to prove that ADEF is a parallelogram would be to do a lot of analytical geometry calculations from the coordinates of the vertices, calculating midpoints fist. From there you would calculate segment lengths, and/or slopes.
I believe that the easier of the many analytical geometry options would be to show that AF and DE are parallel and have the same length. That makes ADEF a parallelogram. (Other combinations of proving segments are congruent and/or parallel seem to involve more calculations).
The midpoint coordinates are the average of the coordinates for the segment ends, so the coordinates are
for D ,
for E ,
for F ,
So segments DE and AF are part of horizontal lines and , and therefore are parallel.
The segments are also congruent, the length of each segment being calculated as the difference of x-coordinates of their endpoints:
DE = and
AF =
A pair of congruent and parallel sides (DE and AF) is enough to prove ADEF is a parallelogram.
Then you can calculate the length of AD or EF, and show that it is not also 6, to prove that ADEF is not a rhombus.
EF =

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Recall that S = side, A = angle.

SAS congruency is a postulate that says that if two triangles have two congruent side lengths, and the angle between them is congruent, then the two triangles are congruent (e.g. if two triangles each have sides of 5 and 6 and the included angle is 57 degrees, then the triangles are congruent).

AAS and ASA are similar; if two triangles have two sets of congruent angles, and the corresponding sides are equal, then the two triangles are congruent.

SSS states that if two triangles have corresponding side lengths (e.g. two 3-4-5 triangles) then the triangles are congruent.

Note that these postulates can also be used to prove similarity.

http://en.wikipedia.org/wiki/Congruence_(geometry)

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so n=90 why
----
Why what?

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why are ad and bc perpendicular to ab and dc?
-------
Maybe they're not.

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2 triangles are formed.
they are triangle ABD and CBD.
these triangles are congruent because:
angle ABD and angle CBD are congruent as a result of the bisection.
BA is congruent to BC as given.
BD is congruent to BD as identity / reflection.
triangles are congruent by SAS
angle ADB is congruent to angle BDC by corresponding parts of congruent triangles.
angle ADC equals sum of angle ADB and angle CDB by adjacent angle sum.
BD bisects angle ADC by separating it into 2 equal angles adjacent to each other.
i don't quite know the exact words to use but this is the general idea on how to prove it.
draw yourself a picture and you will see what i mean.
here's my picture:

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Without a sketch, this is what I am seeing:

1. PA=PB given

2. QA=QB given

3. PQ=PQ reflexive property

4. triangle PAQ = triangle PBQ Side-Side-Side Theorem (SSS)

5. angle APQ = angle BPQ Corresponding parts of congruent triangles are congruent (CPCTC)

6. PR=PR reflexive property

7. triangle APR = triangle BPR Side-Angle-Side Theroem (SAS)

For the final step, we used steps 1, 5, and 6 in order to use the SAS Theorem.

and of course all the equal signs (=) above should be congruency signs.

This is what I'm seeing without looking at your sketch.

You were close on some of the things, but don't forget to use your CPCTC, SAS, SSS, AAS, and ASA Theorems properly!

Also, number your proof steps as I did above, and it will make it easier to read, and easier to get help with.

I hope this helps! :)

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