look at this diagram:
diagram of two parallel vertical lines (ce and fh) cut by a transversal (bi), with points d on ce, g on fh, e above d, c below d, h above g, f below g, b left of d, i right of g
if \overleftrightarrow{ce} and \overleftrightarrow{fh} are parallel lines and m\angle cdg = 70°, what is m\angle cdb?

Question

Accepted Answer

# Explanation:
## Step1: Identify Angle Relationship
Since $ \overleftrightarrow{CE} \parallel \overleftrightarrow{FH} $ and $ \overleftrightarrow{BI} $ is a transversal, $ \angle CDB $ and $ \angle CDG $ are supplementary (they form a linear pair, or consecutive interior angles? Wait, actually, $ \angle CDB $ and $ \angle CDG $ are adjacent and form a straight line? Wait, no, $ \angle CDG = 70^\circ $, and $ \angle CDB $ and $ \angle CDG $ are supplementary because they are adjacent angles on a straight line (linear pair). Wait, no, looking at the diagram: $ D $ is on $ CE $, $ G $ is on $ FH $, and $ BI $ is a transversal. Wait, actually, $ \angle CDB $ and $ \angle HGD $ would be corresponding, but no—wait, $ \angle CDG $ is given as $ 70^\circ $, and $ \angle CDB $ and $ \angle CDG $ are supplementary because they are adjacent and form a linear pair (they add up to $ 180^\circ $)? Wait, no, maybe I misread. Wait, the problem says $ m\angle CDG = 70^\circ $, and we need $ m\angle CDB $. Wait, $ D $ is the intersection of $ BI $ and $ CE $, $ G $ is the intersection of $ BI $ and $ FH $. Since $ CE \parallel FH $, $ \angle CDB $ and $ \angle HGI $ or something? Wait, no, let's think again. The straight line $ BI $ intersects $ CE $ at $ D $ and $ FH $ at $ G $. So $ \angle CDB $ and $ \angle CDG $ are adjacent angles forming a linear pair, so they should add up to $ 180^\circ $. Wait, that makes sense. So if $ m\angle CDG = 70^\circ $, then $ m\angle CDB = 180^\circ - 70^\circ = 110^\circ $? Wait, no, maybe I got the angles wrong. Wait, $ \angle CDB $ and $ \angle CDG $: let's label the points. $ CE $ is a vertical line (up and down), $ FH $ is parallel to $ CE $. $ BI $ is a transversal going from $ B $ (left) through $ D $ (on $ CE $) to $ G $ (on $ FH $) to $ I $ (right). So $ \angle CDG $ is the angle at $ D $ between $ CD $ (down along $ CE $) and $ DG $ (right along $ BI $). $ \angle CDB $ is the angle at $ D $ between $ CD $ (down along $ CE $) and $ DB $ (left along $ BI $). So $ \angle CDB $ and $ \angle CDG $ are adjacent and form a linear pair, so they are supplementary. Therefore, $ m\angle CDB + m\angle CDG = 180^\circ $. So $ m\angle CDB = 180^\circ - 70^\circ = 110^\circ $. Wait, but maybe they are alternate interior angles? No, because $ CE \parallel FH $, and $ BI $ is a transversal. Wait, no, $ D $ is on $ CE $, $ G $ is on $ FH $. So $ \angle CDB $ and $ \angle HGD $ would be corresponding angles, but $ \angle CDG $ is $ 70^\circ $, so $ \angle HGD = 70^\circ $ (vertical angles or corresponding). Then $ \angle CDB $ would be supplementary to $ \angle HGD $ because $ CE \parallel FH $, so consecutive interior angles? Wait, maybe I confused the angles. Let's start over.

## Step1: Recognize Linear Pair
$ \angle CDB $ and $ \angle CDG $ form a linear pair (they are adjacent angles on a straight line $ BI $), so their measures sum to $ 180^\circ $.

## Step2: Calculate $ m\angle CDB $
Given $ m\angle CDG = 70^\circ $, we use the linear pair property:
$ m\angle CDB + m\angle CDG = 180^\circ $
Substitute $ m\angle CDG = 70^\circ $:
$ m\angle CDB = 180^\circ - 70^\circ = 110^\circ $

# Answer:
$ 110^\circ $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Identify Angle Relationship

Step1: Recognize Linear Pair

Step2: Calculate \( m\angle CDB \)

Answer: