Math Final Proofs

Question 1

The Linear Pairs Theorem
Given -∠ACD and ∠DCB from a ln. pr.
Prove - ∠ACD and ∠DCB are supp.

Answer 1

By the SAP rays AC and CB form a line. Therefore by the df. of straight ang., ∠ACB is straight. Then, by STRAP ∠ACB = 180 degrees. Now, ray CD goes through ∠ACB. By the APP m∠ACD+m∠DCB = m∠ACB. By TOE m∠ACD+m∠DCH=180 degrees. Finally concluding that ∠ACD and ∠DCB are supp.

Question 2

The T Theorem

Question 3

The Same Supplements Theorem
Given -angles 1 and 2 are each supplementary to angle 3
Prove - angles 1 and 2 are equal

Answer 3

By the definition of supp.
m∡1 + m∡3 = 180° and m∡2 + m∡3 = 180°. Then you have to subtract then substitute in getting m∡1 = 180° − m∡3 and
m∡2 = 180° − m∡3. Now by TOE m∡1 = m∡2.

Question 4

The Vertical Angles Theorem
Given - angles CBA and DBE are vertical
Prove - m∡CBA = m∡DBE

Answer 4

Ray BA passes through the interior of straight angle CBD,
ray BD passes through the interior of straight angle ABE, by the definition of vertical. Now angles CBA and ABD form a linear pair, angles ABD and DBE form a linear pair by definition of ln. pr. By the LPT angles CBA and ABD are supp, as are ABD and DBE. Therefore m∡CBA = m∡DBE by the SST.

Question 5

The Perpendicular Lines Theorem

Question 6

The Isosceles Triangle Theorem
Given - △ABC with AB=AC
Prove -m∠B=m∠C

Answer 6

First you must construct angle bisector AD of angle ABC. Then by df. of angle bisector m∠BAD=m∠CAD. BY reflexivity AD=AD. Now △BAD≅△CAD from SAS. Lastly m∠B=m∠C by CPCTE.

Question 7

The Equilateral Triangle Theorem
Given - Equilateral triangle ABC
Prove - m∠A=m∠B=m∠C

Answer 7

By the ITT sinces AB=BC, m∠a=m∠C. Again by the ITT since AB=AC, m∠B=m∠C. Finally by TOE, m∠A=m∠B. thus is m∠A=m∠C, and m∠B=m∠C then m∠A=m∠B=m∠C.

Question 8

Side-Side-Side Triangle Congruence
Given - △AMC≅△RST
Prove - △ABC≅△RST

Answer 8

By CPCTE MC=ST. By substitution BC=MC. Then, by the ITT m∠AMB=m∠ABM. Next, by AAP m∠CMB=m∠CBM. Therefore, by EPEW m∠AMC=m∠ABC. By SAS △ABC≅△AMC. Finally, by substitution △ABC≅△RST.

Question 9

Angle-Side-Angle Triangle Congruence

Question 10

The Triangle Exterior Angle Inequality
Given - △ABC
Prove -m∠BCD > m∠B

Answer 10

Construct md. pt. of BC. Call it M. Construct segment AZ through M. such that AM=MZ. Connect Z to C.
We know that m∠BMA=m∠ZMC by the VAT. △BMA≅△CMZ by SAS. By CPCTE m∠ABC=m∠BCZ. Segment CZ bisects angle BCD. Therefore m∠BCZ is less than m∠BCD by whole greater than part. Based on previous information if m∠ABC=BCZ and m∠BCZ<m∠BCD it is only possible that m∠ABC<m∠BCD.

Question 11

CAT
Given - Angles 1 and 2 are corresponding and are equal
Prove - Lines m and n are parallel

Answer 11

Angle 3 is vertical to angle 1. By the VAT angles 1 and 3 are equal. Btu 1 and 2 are also equal. thus by TOE angles 2 and 3 are equal. The AIAT then implies that lines m and n are parallel.

Question 12

AEAT

Question 13

CIAT
Given - Given: transversal t cuts lines a and b, consecutive interior angles 1 and 2 are supplementary
Prove - lines a and b are parallel

Answer 13

Note that angles 3 has been labeled so that angles 2 and 3 are alternate interior.
Now, by the LPT angles 1 and 3 are supplementary. But we were given that angles 1 and 2 are supplementary. Thus by the
SST, angles 2 and 3 are equal. It then follows by the AIAT that lines a and b are parallel.

Question 14

PLT

Question 15

AEAC

Question 16

CIAC
Given - transversal t cuts lines a and b to form consecutive interior angles 1 and 2,
lines a and b are parallel
Prove - angles 1 and 2 are supplementary

Answer 16

Note that as above angles 3 has been labeled so that angles 2 and 3 are alternate interior.
Now, by the LPT angles 1 and 3 are supplementary. Moreover by the AIAC angles 2 and 3 are equal. Thus by substitution
angles 1 and 2 are supplementary.

Question 17

CAC

Question 18

PLC

Question 19

The Triangle Angle Sum Theorem

Question 20

The Third Angle Theorem

Question 21

The Triangle Exterior Angle Equality
Given - △ABC with exterior angle BCD
Prove -

Question 22

The Interior Angle Sum Theorem

Question 23

The Exterior Angle Sum Theorem

Question 24

Angle-Angle-Side Triangle Congruence
Given - △ABC and △RST with m∠A=m∠R, m∠C=m∠T, BC=ST
Prove - △ABC≅ △RST

Answer 24

∠B and ∠S is equal to ∠A=∠C=∠T=∠R by TASTee. Therefore by AAS △ABC≅ △RST

Question 25

Hypotenuse-Leg Triangle Congruence
Given - Right triangle ABC and DEF, right angles C and F, AB = DE, BC=EF
Prove -△ABC≅△DEF

Answer 25

Extend AC to past point C to a point z so that CZ=DF. Connect C to B. Triangle two is congruent to three by SAS. By CPCTE ED=BZ. Triangle one is isosceles. Therefore m∡A=m∡D=m∡Z. By AAS △ABC≅ △DEF.

Question 26

The Converse of the Isosceles Triangle Theorem
Given - △ABC with m∡A=m∡C
Prove - AB=BC

Answer 26

Fist construct angle bisector of ∡ABC. Connect point D. This is now called segment BD. Then use AAS △BDC≅△BDA. Finally by CPCTE AB=BC.

Question 27

The Equiangular Triangle Theorem
Given - △ABC with m∡A=m∡B=m∡C
Prove - AB=BC=CA

Answer 27

By the converse ITT sides CA and BA are equal. Then by the Converse ITT CB is also equal to CA and BA. Therefore AB=BC=CA.

Question 28

The Anisosceles Triangle Theorem
Given: AB > BC
Prove: m∡BCA > m∡BAC

Answer 28

First, we mark the point D on AB such that BD = BC. We then connect D to C. The measure of angle BDC and is greater than that of angle BAC by the TEAI
The measures of angles BDC and BCD are equal by the ITT.
Thus, by substitution the measure of angle BCD is greater than the measure of angle BAC.
The measures of angle BCA is greater than the measure of angle BCD, because the whole is greater than the part.
Finally, by the transitivity of the greater than relation, the measure of angle BCA is greater than the measure of angle
BAC.

Question 29

The Converse of the Anisosceles Triangle Theorem
Given: m∡C > m∡A
Prove: AB > BC

Answer 29

Assumption: AB is not greater than BC.
Reduction: Since AB is not greater than BC, AB = BC or AB < BC. If AB = BC, then angles A and C are equal in measure
by the ITT. That contradicts the Given. If AB < BC, then by the ATT, angle C is less in measure than angle A. But that too
contradicts the Given.

Question 30

The Right Triangle Inequality

Question 31

The Triangle Inequality