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Question

A triangle has vertices labeled upper A, upper B, and upper D. A point on the line segment formed between points upper B and upper D is labeled upper C . A line connects points upper A and upper C, and a ray extends rightward from this line.

Anika constructed ray AC that bisected ∠A . If the m∠BAD is 46 degrees, what is the m∠BAC ?

(1 point)
Responses

The m∠BAC=m∠DAC.
The m angle upper B upper A upper C equals m angle upper D upper A upper C .

The m∠BAC is 46 degrees.
The m angle upper B upper A upper C is 46 degrees.

The m∠BAC is 92 degrees.
The m angle upper B upper A upper C is 92 degrees.

The m∠BAC is 23 degree

Answer 1

Since ray AC bisects angle A, it means that angle BAC and angle DAC are equal. Given that \( m\angle BAD = 46^\circ \), we can express \( m\angle A \) as follows:

\[ m\angle A = m\angle BAC + m\angle DAC \]

Since \( m\angle BAC = m\angle DAC \), let \( m\angle BAC = x \). Thus, we have:

\[ m\angle BAD + m\angle DAC = m\angle A \]

Substituting these values gives us:

\[ 46^\circ + x = 2x \]

Solving for \( x \) gives:

\[ 46^\circ = 2x - x \] \[ 46^\circ = x \]

Thus, the measure of angle BAC is:

\[ m\angle BAC = 46^\circ \]

Therefore, the correct response is:

The \( m\angle BAC \) is 46 degrees.