Mathisconfusing101 said:
Given: XA @ XB; YA @ YB
Prove: XO ^ AB
Statements:
1. XA (similar) XB
2. Arc AX (similar)Arc XB
3. X bisects Arc AB
4. YA (similar) YB
5. Y bisects AB
6. XO (perpendicular) AB
(sry i dunno how to get the symbols... )
um.. so confused... what do i use to "prove" and what am I proving?
Mathisconfusing101, what you post here is confusing to US, too.
If you are not sure how to post symbols, use words instead.
I've dusted off my crystal ball, and it tells me that you are
GIVEN: XA is congruent to XB
and, YA is congruent to YB
PROVE: XO is perpendicular to AB
I'd do the proof this way (since I can make NO sense of what you have posted):
XA is congruent to XB Given
YA is congurent to YB Given
XY is congruent to Xy Reflexive property of congruence (any segment is congruent to itself)
Triangle AYX is congruent to triangle BYX SSS congruence postulate
<AYX is congruent to <BYX Corresponding angles of congruent triangles are congruent
m < AYX = m <BYX Congruent angles have the same measure
m < AYX + m < BYX = m <AYB Angle addition postulate
m < AYB = 180 definition of a straight angle
m <AYX + m<BYX = 180 Substitution
m <AYX + m<AYX = 180 Subsititution of m <AYX for m <BYX
2 m<AYX = 180 Arithmetic (adding like terms)
m <AYX = 90 Division property of equality (divide both sides by 2)
<AYX is a right angle Definition of a right angle
AB is perpendicular to XO Definition of perpendicular (if two lines intersect to form a right angle, then the lines are perpendicular).
See....I did not use any special symbols here.
Also, I may have misinterpreted your problem...since I had to GUESS at what you meant.
