SOLUTION: Given: AX || CY
BA bisects ∠CAX,
BC bisects ∠ACY.
Prove: ∠B is a right angle.
Diagram looks like a triangle (where it's pointed end is pointing in the righ
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Question 762959: Given: AX || CY
BA bisects ∠CAX,
BC bisects ∠ACY.
Prove: ∠B is a right angle.
Diagram looks like a triangle (where it's pointed end is pointing in the right) in the middle of the parallel lines.
Looks something like this:
A
__________X
|
| .C
|_________Y
B
Has to be written as Statements & Reasons. I really want to know how to do this!
Answer by ramkikk66(644) (Show Source): You can put this solution on YOUR website!
See the picture below which describes your problem. I have extended line CA further upwards to a point P, which is required for the proof.
Problem:
AX || CY
BA bisects ?CAX,
BC bisects ?ACY.
Prove: ?B is a right angle.
Step 1:
BA bisects angle CAX. So angle CAB = angle BAX
Step 2:
BC bisects angle ACY. So angle ACB = angle BCY
Step 3:
Angle PAX = angle ACY since AX || CY (external angles of parallel lines)
Step 4:
Angles PAX + CAX = 180 (supplementary angles on a straight line)
Step 5:
i.e. PAX / 2 + CAX / 2 = 90
Step 6:
ACY / 2 + CAX / 2 = 90 Since PAX = ACY
Step 7:
ACB + CAB = 90 Since ACB is half of ACY, CAB is half of CAX
Step 8:
But in Triangle ABC, ACB + CAB + ABC = 180
Step 9:
So ABC = Angle B = 90
:)
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