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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 is 23 degrees.
The m angle upper B upper A upper C is 23 degrees.

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.

Answer 1

Given that ray AC bisects angle A, it means that the measure of angle BAC is equal to the measure of angle CAD.

Since it is given that \( m\angle BAD = 46^\circ \), we can express angle BAD as the sum of angles BAC and CAD:

\[ m\angle BAD = m\angle BAC + m\angle CAD \]

Since ray AC bisects angle A, we have:

\[ m\angle BAC = m\angle CAD \]

Let's denote \( m\angle BAC = x \). Then:

\[ m\angle BAD = x + x = 2x \]

Setting it equal to 46 degrees gives us:

\[ 2x = 46^\circ \]

Solving for \( x \):

\[ x = \frac{46^\circ}{2} = 23^\circ \]

Thus, \( m\angle BAC = 23^\circ \).

So the correct response is:

The \( m\angle BAC \) is 23 degrees.