Question

19 multiple choice 1 point
in the figure shown, line segments ad and bc are parallel. what is the value of x?
70
110
80
60

Explanation:

Step1: Find angle at F in triangle DFC

In triangle \( DFC \), sum of angles is \( 180^\circ \). So \( \angle DFC = 180^\circ - 70^\circ - 60^\circ = 50^\circ \)? Wait, no, wait. Wait, \( \angle BFG \) and \( \angle DFC \) are vertical angles, so they are equal. Wait, first, let's look at triangle \( BFG \): angles are \( 60^\circ \), \( 60^\circ \), so \( \angle BFG = 180 - 60 - 60 = 60^\circ \)? Wait, no, maybe I messed up. Wait, the problem says \( AD \parallel BC \). Let's find \( \angle FBC \) or \( \angle FCB \)? Wait, no, let's find the angle at \( F \) first. Wait, in triangle \( DFC \), angles are \( \angle D = 70^\circ \), \( \angle C = 60^\circ \), so \( \angle DFC = 180 - 70 - 60 = 50^\circ \)? No, that can't be. Wait, no, \( \angle BFG \) and \( \angle DFC \) are vertical angles, so \( \angle BFG = \angle DFC \). Then in triangle \( BFG \), angles are \( \angle G = 60^\circ \), \( \angle BFG = \angle DFC \), and \( \angle FBG \). Wait, maybe better to find \( \angle FBC \). Wait, since \( AD \parallel BC \), \( \angle ADB = \angle DBC \) (alternate interior angles). Wait, \( \angle ADB \) is \( 70^\circ \)? No, \( \angle ADC = 70^\circ \), and \( AD \parallel BC \), so \( \angle ADC + \angle BCD = 180^\circ \)? No, \( AD \parallel BC \), so \( \angle ADB = \angle DBC \). Wait, maybe another approach. Let's find the angle at \( F \) in triangle \( BFG \). Wait, triangle \( BFG \): \( \angle G = 60^\circ \), \( \angle BFG \) is vertical to \( \angle DFC \). Wait, in triangle \( DFC \), \( \angle D = 70^\circ \), \( \angle C = 60^\circ \), so \( \angle DFC = 180 - 70 - 60 = 50^\circ \)? No, that's 50? Wait, 70 + 60 is 130, 180 - 130 is 50. Then \( \angle BFG = 50^\circ \) (vertical angles). Then in triangle \( BFG \), angles are \( 60^\circ \) (at G), \( 50^\circ \) (at F), so \( \angle FBG = 180 - 60 - 50 = 70^\circ \)? Wait, no, that doesn't make sense. Wait, maybe I made a mistake. Wait, the problem says \( AD \parallel BC \), so \( \angle A + \angle ABC = 180^\circ \) (if \( AB \) is a transversal). Wait, maybe the quadrilateral \( ABCD \) has \( AD \parallel BC \), so it's a trapezoid. Wait, let's find \( \angle ABC \). Wait, \( \angle DFC = 180 - 70 - 60 = 50^\circ \), so \( \angle BFG = 50^\circ \). Then in triangle \( BFG \), \( \angle G = 60^\circ \), \( \angle BFG = 50^\circ \), so \( \angle FBG = 180 - 60 - 50 = 70^\circ \). Then, since \( AD \parallel BC \), \( \angle A + \angle ABC = 180^\circ \). Wait, \( \angle ABC = \angle FBG \)? No, \( \angle ABC \) is \( \angle FBG \) plus something? Wait, maybe I messed up. Wait, let's start over.

Wait, the key is that \( AD \parallel BC \), so \( \angle A + \angle ABC = 180^\circ \) (consecutive interior angles). Now, let's find \( \angle ABC \). First, find \( \angle FBC \). In triangle \( DFC \), angles are \( 70^\circ \), \( 60^\circ \), so \( \angle DFC = 180 - 70 - 60 = 50^\circ \). Then \( \angle BFG = 50^\circ \) (vertical angles). In triangle \( BFG \), angles are \( 60^\circ \) (at G), \( 50^\circ \) (at F), so \( \angle FBG = 180 - 60 - 50 = 70^\circ \). Wait, no, that's not right. Wait, triangle \( BFG \): \( \angle G = 60^\circ \), \( \angle B = \)? Wait, maybe the triangle \( BFG \) has two angles of \( 60^\circ \), so it's isoceles? Wait, the diagram shows \( \angle G = 60^\circ \), and maybe \( \angle FBG = 60^\circ \), so \( \angle BFG = 60^\circ \). Then \( \angle DFC = 60^\circ \) (vertical angles). Then in triangle \( DFC \), \( \angle D = 70^\circ \), \( \angle C = 60^\circ \), \( \angle DFC = 60^\circ \)? But 70 + 60 + 60 = 190, which is more than 180. So that's wrong. Wait, I must have misread the diagram. Let me re-examine. The diagram: points A, B, G; D, F, C. So \( AD \) and \( BC \) are parallel. \( \angle D = 70^\circ \), \( \angle C = 60^\circ \), \( \angle G = 60^\circ \). So maybe \( \angle FDC = 70^\circ \), \( \angle FCD = 60^\circ \), so \( \angle DFC = 180 - 70 - 60 = 50^\circ \). Then \( \angle BFG = 50^\circ \). In triangle \( BFG \), \( \angle G = 60^\circ \), \( \angle BFG = 50^\circ \), so \( \angle FBG = 180 - 60 - 50 = 70^\circ \). Now, since \( AD \parallel BC \), \( \angle A + \angle FBG = 180^\circ \)? No, wait, \( AD \parallel BC \), so \( \angle A + \angle ABC = 180^\circ \). Wait, \( \angle ABC \) is \( \angle FBG \)? No, \( \angle ABC \) is \( \angle FBC \), which is equal to \( \angle FBG \)? Wait, maybe \( AB \parallel DG \)? No, the problem says \( AD \parallel BC \). Wait, maybe the quadrilateral \( ABGD \) is a parallelogram? No, not necessarily. Wait, another approach: the sum of angles in a triangle is 180. Let's find the angle at \( F \) in triangle \( BFG \). Wait, maybe the angle at \( F \) is equal to the angle at \( D \) or something. Wait, I think I made a mistake. Let's try again.

Wait, the correct approach: since \( AD \parallel BC \), \( \angle ADB = \angle DBC \) (alternate interior angles). Wait, \( \angle ADB \) is \( 70^\circ \), so \( \angle DBC = 70^\circ \). Then in triangle \( BFG \), \( \angle G = 60^\circ \), \( \angle BFG = \angle DFC \). Wait, no, let's find \( \angle BFC \). Wait, maybe the answer is 50? No, the options are 70, 110, 80, 60. Wait, maybe I messed up the triangle. Wait, let's calculate the angle at \( F \) in triangle \( DFC \): \( 180 - 70 - 60 = 50 \). Then \( \angle BFG = 50 \). Then in triangle \( BFG \), \( \angle G = 60 \), \( \angle BFG = 50 \), so \( \angle FBG = 70 \). Then since \( AD \parallel BC \), \( \angle A + \angle FBG = 180 \)? No, that would make \( \angle A = 110 \). Ah! That's it. So \( \angle A + \angle FBG = 180^\circ \) because \( AD \parallel BC \) and \( AB \) is a transversal. So \( x + 70^\circ = 180^\circ \)? Wait, no, \( \angle FBG \) is \( 70^\circ \)? Wait, no, if \( \angle FBG = 70^\circ \), then \( x = 180 - 70 = 110 \). Yes, that makes sense. So the value of \( x \) is 110.

Step1: Calculate \( \angle DFC \) in \( \triangle DFC \)

In \( \triangle DFC \), sum of angles is \( 180^\circ \).
\( \angle DFC = 180^\circ - 70^\circ - 60^\circ = 50^\circ \).

Step2: Find \( \angle FBG \) in \( \triangle BFG \)

\( \angle BFG = \angle DFC = 50^\circ \) (vertical angles).
In \( \triangle BFG \), sum of angles is \( 180^\circ \):
\( \angle FBG = 180^\circ - 60^\circ - 50^\circ = 70^\circ \).

Step3: Use \( AD \parallel BC \) to find \( x \)

Since \( AD \parallel BC \), \( \angle A + \angle FBG = 180^\circ \) (consecutive interior angles).
\( x + 70^\circ = 180^\circ \)
\( x = 180^\circ - 70^\circ = 110^\circ \).

Answer:

110