Question

if $overrightarrow{bd}$ bisects $angle abc$ and $mangle 3 = 80$, what is $mangle abd$?
$mangle abd = \square$

Explanation:

Step1: Identify vertical angles or linear pairs

We know that ∠2 and ∠3 are adjacent and form a linear pair? Wait, no, first, ∠2 and ∠3: Wait, actually, ∠2 and ∠3 are adjacent? Wait, no, looking at the diagram, the two vertical lines (one with 1,2,3,4 and the other with A and the vertical arrow) intersect at B? Wait, no, the line with 1,2,3,4 intersects the line with A and the vertical arrow at B. So ∠3 and ∠2: Wait, ∠3 and ∠2 are adjacent? Wait, no, ∠2 and ∠3 are supplementary? Wait, no, first, ∠3 is given as 80 degrees. Wait, the line with 1,2,3,4: ∠2 and ∠3 are adjacent, and ∠2 + ∠3 = 180? Wait, no, if the two lines are intersecting, then ∠2 and ∠3 are adjacent and form a linear pair? Wait, no, the angle at B: the line with A is vertical, and the other line (with 1,2,3,4) is a transversal? Wait, maybe ∠ABC is a straight angle? Wait, BD bisects ∠ABC. So first, let's find ∠ABC.

Wait, the line with 1,2,3,4: ∠3 is 80 degrees. The angle adjacent to ∠3 (∠2) and ∠3: Wait, ∠2 and ∠3 are supplementary? Wait, no, if the two lines are intersecting, then ∠2 and ∠3 are adjacent and form a linear pair, so ∠2 + ∠3 = 180? Wait, no, maybe ∠ABC is equal to ∠2? Wait, no, the line with A is vertical, and the other line (with 1,2,3,4) is a transversal. Wait, maybe ∠ABC is a straight angle? Wait, BD bisects ∠ABC, so ∠ABD = ∠DBC.

Wait, first, let's find ∠ABC. The angle ∠3 is 80 degrees, and the angle ∠2 is vertical to some angle? Wait, no, the line with 1,2,3,4: ∠3 and ∠1 are vertical angles? Wait, no, ∠1 and ∠3 are vertical? Wait, no, ∠1 and ∠3 are vertical angles? Wait, ∠1 and ∠3: when two lines intersect, vertical angles are equal. Wait, the two lines (the one with 1,2,3,4 and the one with A and the vertical arrow) intersect at B. So ∠3 and ∠1 are vertical? No, ∠2 and ∠4 are vertical, ∠1 and ∠3 are vertical? Wait, no, ∠1 and ∠3: if the lines are intersecting, then ∠1 and ∠3 are vertical angles, so they are equal. Wait, but ∠3 is 80 degrees, so ∠1 is 80? No, that can't be. Wait, maybe ∠ABC is a straight angle, so ∠ABC = 180 - ∠3? Wait, no, let's re-examine.

Wait, the diagram: there's a line with points 1,2,3,4 (a transversal) intersecting a vertical line (with A and the vertical arrow) at B. Then, from B, there are two rays: BD and BC. BD bisects ∠ABC. So ∠ABC is the angle between BA (vertical) and BC. Wait, the transversal line (with 1,2,3,4) intersects the vertical line at B, so ∠3 is adjacent to ∠ABC? Wait, ∠3 and ∠ABC: are they supplementary? Wait, ∠3 is 80 degrees, so ∠ABC = 180 - 80 = 100? No, that doesn't make sense. Wait, maybe ∠ABC is equal to ∠2? Wait, ∠2 and ∠3 are supplementary, so ∠2 = 180 - 80 = 100. Then ∠ABC is equal to ∠2, so ∠ABC = 100 degrees? Wait, no, maybe ∠ABC is a straight angle? Wait, no, BD bisects ∠ABC, so ∠ABD = ∠DBC = 1/2 ∠ABC.

Wait, let's start over. The problem says BD bisects ∠ABC, so ∠ABD = ∠DBC. We need to find ∠ABD. First, find ∠ABC. The angle ∠3 is 80 degrees. Looking at the diagram, ∠3 and ∠ABC: are they vertical angles? No, ∠3 and ∠ABC: wait, the line with 1,2,3,4 and the line with A and the vertical arrow intersect at B, so ∠3 and ∠ABC: maybe ∠ABC is equal to ∠2, and ∠2 and ∠3 are supplementary. So ∠2 + ∠3 = 180, so ∠2 = 180 - 80 = 100. Then ∠ABC = ∠2 = 100 degrees? Wait, no, maybe ∠ABC is a straight angle? Wait, no, if BD bisects ∠ABC, then ∠ABC is the angle between BA and BC, and BD is the bisector. So if ∠ABC is 100 degrees, then ∠ABD is 50? No, that doesn't fit. Wait, maybe ∠3 and ∠ABC are vertical angles? No, ∠3 is 80, so ∠ABC is 80? Then BD bisects it, so ∠ABD is 40? No, that also doesn't fit. Wait, maybe I'm misinterpreting the diagram.

Wait, the diagram: there's a vertical line (with A at the top, B in the middle, and a downward arrow). Then a transversal line (with 1,2,3,4) intersects the vertical line at B. So ∠3 is the angle between the transversal (going to the right, downwards) and the vertical line (downwards from B). Then, from B, there are two rays: BD (going to the right, upwards) and BC (going to the right, downwards). So ∠ABC is the angle between BA (upwards vertical) and BC (downwards right). Wait, no, BA is upwards, BC is downwards right? No, the diagram shows A above B, and C below B? Wait, no, the arrows: A is above B, with an upward arrow. The transversal line has 1,2,3,4: 1 is top left, 2 is top right (between transversal and vertical line), 3 is bottom right (between transversal and vertical line), 4 is bottom left. Then BD is a ray from B to D (right, upwards), and BC is a ray from B to C (right, downwards). So ∠ABC is the angle between BA (upwards) and BC (downwards right). The transversal line makes ∠3 with the vertical line (downwards from B). So ∠3 is 80 degrees, so the angle between the transversal (downwards right) and the vertical line (downwards) is 80 degrees. Then the angle between the transversal (upwards left) and the vertical line (upwards) is also 80 degrees (vertical angles). Wait, no, ∠2 is the angle between transversal (upwards left) and vertical line (upwards), so ∠2 = 80? No, ∠2 and ∠3 are supplementary? Wait, ∠2 + ∠3 = 180, so ∠2 = 100. Then ∠ABC is equal to ∠2? Wait, ∠ABC is the angle between BA (upwards) and BC (downwards right). The transversal line (upwards left) and BA (upwards) form ∠2, and the transversal line (downwards right) and BC (downwards right) form ∠3. Wait, maybe ∠ABC is a straight angle? No, BA and BC are not a straight line. Wait, maybe BD bisects ∠ABC, so ∠ABD = ∠DBC. And ∠ABC is equal to ∠2, which is 100 degrees? Then ∠ABD would be 50? No, that doesn't make sense. Wait, maybe ∠3 and ∠ABC are vertical angles? No, ∠3 is 80, so ∠ABC is 80. Then BD bisects it, so ∠ABD is 40? No, that also doesn't fit. Wait, maybe I made a mistake.

Wait, let's think again. The problem says BD bisects ∠ABC, so ∠ABD = ∠DBC. We need to find ∠ABD. First, find ∠ABC. The angle ∠3 is 80 degrees. Looking at the diagram, ∠3 and ∠ABC: are they adjacent? Wait, the transversal line and the line BC: maybe ∠3 and ∠DBC are vertical angles? No, BD is a bisector. Wait, maybe ∠ABC is a straight angle, so ∠ABC = 180 - ∠3? Wait, ∠3 is 80, so ∠ABC = 100? Then BD bisects it, so ∠ABD = 50? No, that's not right. Wait, maybe ∠3 and ∠ABC are equal? So ∠ABC = 80, then ∠ABD = 40? No, that's not. Wait, maybe the diagram is such that ∠ABC is a straight angle, so ∠ABC = 180 degrees, and ∠3 is 80, so the angle adjacent to ∠3 is 100, but BD bisects ∠ABC, so ∠ABD = 90? No, that's not. Wait, I think I'm overcomplicating. Let's recall that when a line bisects an angle, it divides it into two equal parts. So if we can find ∠ABC, then divide by 2.

Looking at the diagram, the line with 1,2,3,4 intersects the vertical line at B. So ∠3 and ∠2 are supplementary (they form a linear pair), so ∠2 + ∠3 = 180. Since ∠3 = 80, ∠2 = 180 - 80 = 100. Now, ∠ABC is equal to ∠2 (vertical angles or corresponding angles? Wait, ∠ABC and ∠2: are they vertical angles? Yes, because the two lines (transversal and vertical) intersect at B, so ∠2 and ∠ABC are vertical angles, so they are equal. Therefore, ∠ABC = ∠2 = 100 degrees. Now, BD bisects ∠ABC, so ∠ABD = ∠DBC = 1/2 ∠ABC. Therefore, ∠ABD = 100 / 2 = 50 degrees? Wait, no, that can't be. Wait, maybe ∠ABC is a straight angle, so ∠ABC = 180 degrees, and ∠3 is 80, so the angle between BA and the transversal is 80, so ∠ABC = 180 - 80 = 100? No, I'm confused.

Wait, let's start over. The problem is: BD bisects ∠ABC, m∠3 = 80, find m∠ABD.

From the diagram, ∠3 and ∠ABC: are they adjacent? Wait, the transversal line (with 1,2,3,4) and the line BC: maybe ∠3 and ∠DBC are vertical angles? No, BD is a bisector. Wait, maybe ∠ABC is a straight angle, so ∠ABC = 180 degrees, and ∠3 is 80, so the angle between BA and the transversal is 80, so ∠ABD is 80? No, that's not. Wait, maybe the diagram is such that ∠3 and ∠ABD are vertical angles? No, BD bisects ∠ABC. Wait, maybe ∠ABC is equal to ∠3, so ∠ABC = 80, then ∠ABD = 40? No, that's not. Wait, I think I made a mistake in the angle. Let's look at the diagram again.

The diagram: there's a vertical line (A above B, downward arrow below B). A transversal line crosses the vertical line at B, creating angles 1,2,3,4. Angle 3 is between the transversal (going down to the right) and the vertical line (downward from B). Then, from B, there are two rays: BD (going up to the right) and BC (going down to the right). So ∠ABC is the angle between BA (upward) and BC (downward right). The transversal line (downward right) and BC: are they the same line? No, BC is a ray, and the transversal is a line. Wait, maybe BC is along the transversal? So the transversal line is BC? Then ∠ABC is the angle between BA and BC (the transversal). Then ∠3 is the angle between the transversal (BC) and the vertical line (BA downward). Wait, ∠3 is 80, so the angle between BC (transversal) and BA (downward) is 80, so the angle between BA (upward) and BC (transversal) is 180 - 80 = 100 degrees (since BA is a straight line, upward and downward). So ∠ABC = 100 degrees. Then BD bisects ∠ABC, so ∠ABD = 100 / 2 = 50 degrees? Wait, but that seems off. Wait, no, if BA is a straight line (upward and downward), then the angle between upward BA and downward BA is 180 degrees. The transversal (BC) makes ∠3 = 80 degrees with downward BA, so the angle between upward BA and BC is 180 - 80 = 100 degrees, which is ∠ABC. Then BD bisects ∠ABC, so ∠ABD = 50 degrees. Yes, that makes sense.

Step1: Find ∠ABC

∠3 and ∠ABC are supplementary (since they form a linear pair with the straight line BA). Wait, no, BA is a straight line, so the angle between upward BA and downward BA is 180°. ∠3 is the angle between downward BA and BC (transversal), so ∠ABC (angle between upward BA and BC) is 180° - ∠3.
Given \( m\angle 3 = 80^\circ \),
\( m\angle ABC = 180^\circ - 80^\circ = 100^\circ \).

Step2: Bisect ∠ABC with BD

Since \( \overrightarrow{BD} \) bisects \( \angle ABC \), it divides \( \angle ABC \) into two equal angles: \( \angle ABD \) and \( \angle DBC \).
Thus, \( m\angle ABD = \frac{1}{2} m\angle ABC \).

Substitute \( m\angle ABC = 100^\circ \):
\( m\angle ABD = \frac{1}{2} \times 100^\circ = 50^\circ \).

Answer:

\( 50^\circ \)