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# Proofs and conjectures

## Proofs and conjectures in geometry

You have seen how to use geometry and the properties of polygons to help you find unknown lengths and angles in various quadrilaterals and polygons. We will now extend this work to proving some of the properties and to solving riders. A conjecture is the mathematicians way of saying I believe that this is true, but I have no proof. The following worked examples will help make this clearer.

### Exercise 1: Proofs - 1

Given quadrilateral ABCD, with ABCDABCD and ADBCADBC, prove that B A ^ D = B C ^ A B A ^ D= B C ^ A and A B ^ C = A D ^ C A B ^ C= A D ^ C.

#### Solution

1. Step 1. Draw a diagram: We draw the following diagram and construct the diagonals.
2. Step 2. Write down what you are given and what you need: Given: ABCDABCD and ADBCADBC. We need to prove A=CA=C and B=DB=D. In the formal language of maths we say that we are required to prove (RTP) B A ^ D = B C ^ A B A ^ D= B C ^ A and A B ^ C = A D ^ C A B ^ C= A D ^ C.
3. Step 3. Solve the problem:
B A ^ C = A C ^ D ( corresponding angles ) D A ^ C = B C ^ A ( corresponding angles ) B A ^ D = B C ^ A B A ^ C = A C ^ D ( corresponding angles ) D A ^ C = B C ^ A ( corresponding angles ) B A ^ D = B C ^ A
(1)
Similarly we find that:
A B ^ C = A D ^ C A B ^ C= A D ^ C
(2)

### Proofs - 2 2

In parallelogram ABCD, the bisectors of the angles (AW, BX, CY and DZ) have been constructed:

You are also given that AB=CDAB=CD, AD=BCAD=BC, ABCDABCD, ADBCADBC, A ^ = C ^ A ^ = C ^ , and B ^ = D ^ B ^ = D ^ . Prove that MNOP is a parallelogram.

#### Solution

1. Step 1. Write down what you are given and what you need to prove: Given: AB=CDAB=CD, AD=BCAD=BC, ABCDABCD, ADBCADBC, A ^ = C ^ A ^ = C ^ , and B ^ = D ^ B ^ = D ^ . RTP: MNOP is a parallelogram.
2. Step 2. Solve the problem:
In ADW and CBY D A ^ W = B C ^ Y ( given ) A D ^ C = A B ^ C ( given ) AD = BC(given) ADW = CBY(AAS) DW=BY In ADW and CBY D A ^ W = B C ^ Y ( given ) A D ^ C = A B ^ C ( given ) AD = BC(given) ADW = CBY(AAS) DW=BY
(3)
In ABX and CDZ D C ^ Z = B A ^ X ( given ) Z D ^ C = X B ^ A ( given ) DC = AB(given) ABX CDZ(AAS) AX=CZ In ABX and CDZ D C ^ Z = B A ^ X ( given ) Z D ^ C = X B ^ A ( given ) DC = AB(given) ABX CDZ(AAS) AX=CZ
(4)
In XAM and ZCO X A ^ M = Z C ^ O ( given ) A X ^ M = C Z ^ O ( proven above ) AX = CZ(proven above) XAM COZ(AAS) A O ^ C = A M ^ X In XAM and ZCO X A ^ M = Z C ^ O ( given ) A X ^ M = C Z ^ O ( proven above ) AX = CZ(proven above) XAM COZ(AAS) A O ^ C = A M ^ X
(5)
A M ^ X = P M ^ N(vert. opp. ∠'s) C O ^ Z = N O ^ P(vert. opp. ∠'s) P M ^ N = N O ^ P A M ^ X = P M ^ N(vert. opp. ∠'s) C O ^ Z = N O ^ P(vert. opp. ∠'s) P M ^ N = N O ^ P
(6)
In BYN and DWP Y B ^ N = W D ^ P ( given ) B Y ^ N = W D ^ P ( proven above ) DW = BY(proven above) YBN WDP(AAS) B N ^ Y = D P ^ W In BYN and DWP Y B ^ N = W D ^ P ( given ) B Y ^ N = W D ^ P ( proven above ) DW = BY(proven above) YBN WDP(AAS) B N ^ Y = D P ^ W
(7)
D P ^ W = M P ^ O(vert. opp. ∠'s) B N ^ Y = O N ^ M(vert. opp. ∠'s) M P ^ O = O N ^ M D P ^ W = M P ^ O(vert. opp. ∠'s) B N ^ Y = O N ^ M(vert. opp. ∠'s) M P ^ O = O N ^ M
(8)

MNOP is a parallelogram (both pairs opp. 's ==, and therefore both pairs opp. sides parallel too)

### Warning:

It is very important to note that a single counter example disproves a conjecture. Also numerous specific supporting examples do not prove a conjecture.

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