Did I do this one right?

[Wormman12]

Okay, I tried doing this one, and this is what I came up with:

1. Using what we know of this question:
a. <CAX and <ACY=90°
i. <CAB=1/2 <CAX. <ACB=1/2<ACY
b. <CAB and <ACB=45°
c. The Sum of angles for a triangle = 180°
d. Thus; 45+45+m<B=180; m<B=90.

This is the only way that it makes sense to me, but it doesn't identify the two angles as being right angles in the paper, so I'm worried I have it wrong.View attachment Mathmatical problem1.pdf

 

[DrPhil]

Okay, I tried doing this one, and this is what I came up with:

1. Using what we know of this question:
a. <CAX and <ACY=90° You can NOT assume this!
i. <CAB=1/2 <CAX. <ACB=1/2<ACY
b. <CAB and <ACB=45° Nope.
c. The Sum of angles for a triangle = 180°
d. Thus; 45+45+m<B=180; m<B=90.

This is the only way that it makes sense to me, but it doesn't identify the two angles as being right angles in the paper, so I'm worried I have it wrong.View attachment 2565

Use the theorem about angles made by a line (AC) crossing two parallel lines (AX and CY).
What can you say about angles <CAX and <ACY? In particular, what is their sum?

Carry on!

 

[Wormman12]

Use the theorem about angles made by a line (AC) crossing two parallel lines (AX and CY).
What can you say about angles <CAX and <ACY? In particular, what is their sum?

Carry on!

These angles would be interior angles on the same side of a transversal line. Meaning that their equal sum would b 180 degrees. They are both bisexted, so the sum of angles CAB and ACB would be complimentary. That means they make up 90 degrees of the 180 found in a triangle, and thus angle B is 90 degrees. Is that correct?

 

[DrPhil]

These angles would be interior angles on the same side of a transversal line. Meaning that their equal sum would b 180 degrees. They are both bisexted, so the sum of angles CAB and ACB would be complimentary. That means they make up 90 degrees of the 180 found in a triangle, and thus angle B is 90 degrees. Is that correct?

You got it!

 

[soroban]

Hello, Wormman12!

Let's start with the original problem . . .

\(\displaystyle \text{Given: }\;AX \parallel CY,\;\;BA\text{ bisects }\angle CAX,\;\;BC\text{ bisects } \angle ACY\)
\(\displaystyle \text{Prove: }\;\angle B = 90^o\)

              A *  *  *  *  *  * X
               *  *  α
              *  α  *
             *        *
            *           *
           *              * B
          *           *
         *        *
        * β   *
       *  *  β
    C *  *  *  *  *  *  *  *  * Y

\(\displaystyle BA\text{ bisects }\angle CAX.\)
\(\displaystyle \text{Let }\angle CAB = \angle BAX = \alpha\)
Then \(\displaystyle \angle CAX = 2\alpha\)

\(\displaystyle BC\text{ bisects }\angle ACY.\)
\(\displaystyle \text{Let }\angle ACB = \angle BCY = \beta\)
Then \(\displaystyle \angle ACY = 2\beta\)

\(\displaystyle \text{Since }AX \parallel CY\!:\:\angle CAX + \angle ACY \:=\:180^o\)
\(\displaystyle \text{Hence: }\:2\alpha + 2\beta \:=\:180^o \quad\Rightarrow\quad \alpha + \beta \:=\:90^o \;\;\color{purple}{[1]}\)

\(\displaystyle \text{In }\Delta ABC\!:\;\angle B + \alpha + \beta \:=\:180^o\)

\(\displaystyle \text{Substitute }\color{purple}{[1]}\!:\;\angle B + 90^o \:=\:180^o\)

\(\displaystyle \text{Therefore: }\:\angle B \:=\:90^o\)