select the correct answer.
given: $\overline{ad} \parallel \overline{cb}$, and $\overline{md} \cong \overline{mb}$
prove: $am = cm$
image of a geometric figure with points a, m, c on a horizontal line, and b, m, d on a vertical line, with segments ad and cb having arrows indicating parallelism, and md and mb having tick marks indicating congruence
| statements | reasons |
| --- | --- |
| $\angle dma \cong \angle bmc$ | vertical angles theorem |
| $\overline{md} \cong \overline{mb}$ | given |
| $\overline{ad} \parallel \overline{cb}$ | given |
| $\angle mda \cong \angle mbc$ | alternate interior angles theorem |
| ? | ? |
| $\overline{am} \cong \overline{cm}$ | cpctc |
| $am = cm$ | definition of congruence |
which step is missing in the proof?
$\circ$ a. $\triangle mda \cong \triangle bmc$ by sas
$\circ$ b. $\triangle mda \cong \triangle mbc$ by sas
$\circ$ c. $\triangle mda \cong \triangle mcb$ by asa
$\circ$ d. $\triangle mda \cong \triangle mbc$ by asa

Question

Accepted Answer

# Explanation:
## Step1: Recall SAS and ASA criteria
SAS: Two sides and included angle. ASA: Two angles and included side.
## Step2: Analyze given congruences
We have $ \angle MDA \cong \angle MBC $ (alternate interior), $ \overline{MD} \cong \overline{MB} $ (given), $ \angle DMA \cong \angle BMC $ (vertical angles). Wait, no—wait, the angles around $ M $: $ \angle MDA $ and $ \angle MBC $ (alternate interior), $ \overline{MD} \cong \overline{MB} $, and $ \angle DMA \cong \angle BMC $? Wait, no, let's check the triangles. Triangles $ MDA $ and $ MBC $: $ \angle MDA \cong \angle MBC $ (alternate interior), $ \overline{MD} \cong \overline{MB} $ (given), $ \angle DMA \cong \angle BMC $? No, wait, vertical angles are $ \angle DMA $ and $ \angle BMC $? Wait, no, $ \angle DMA $ and $ \angle BMC $ are vertical angles? Wait, the intersection is at $ M $, so $ \angle DMA $ and $ \angle BMC $ are vertical angles? Wait, no, $ \angle DMA $ and $ \angle BMC $: actually, when $ AD \parallel CB $, and transversal $ BD $, so alternate interior angles are $ \angle MDA $ and $ \angle MBC $. Then, we have $ \overline{MD} \cong \overline{MB} $ (side), $ \angle MDA \cong \angle MBC $ (angle), and $ \angle DMA \cong \angle BMC $ (angle)? Wait, no, the vertical angles are $ \angle DMA $ and $ \angle BMC $? Wait, no, $ \angle DMA $ and $ \angle BMC $: let's see the diagram. Point $ M $ is on $ AC $ and $ BD $. So $ \angle DMA $ and $ \angle BMC $ are vertical angles (since $ BD $ and $ AC $ intersect at $ M $). So we have two angles and a side? Wait, no, the sides: $ MD = MB $ (given), $ \angle MDA = \angle MBC $ (alternate interior), and $ \angle DMA = \angle BMC $ (vertical angles). Wait, but the SAS would be: side $ MD = MB $, angle $ \angle MDA = \angle MBC $, and side $ MA $ and $ MC $? No, wait, no. Wait, the triangles are $ \triangle MDA $ and $ \triangle MBC $. Let's list the parts:

- $ \angle MDA \cong \angle MBC $ (alternate interior angles, since $ AD \parallel CB $ and $ BD $ is transversal)
- $ \overline{MD} \cong \overline{MB} $ (given)
- $ \angle DMA \cong \angle BMC $ (vertical angles theorem)

Wait, that's two angles and a side, but the side is between the angles? Wait, no, $ \overline{MD} $ is between $ \angle MDA $ and $ \angle DMA $, and $ \overline{MB} $ is between $ \angle MBC $ and $ \angle BMC $. So that's ASA? Wait, no, ASA is two angles and the included side. So $ \angle MDA $ (angle), $ \overline{MD} $ (side), $ \angle DMA $ (angle) in $ \triangle MDA $; and $ \angle MBC $ (angle), $ \overline{MB} $ (side), $ \angle BMC $ (angle) in $ \triangle MBC $. Since $ \overline{MD} \cong \overline{MB} $, $ \angle MDA \cong \angle MBC $, $ \angle DMA \cong \angle BMC $, so by ASA, $ \triangle MDA \cong \triangle MBC $. Wait, but the options: option D is $ \triangle MDA \cong \triangle MBC $ by ASA. Wait, but let's check the options:

Option B: $ \triangle MDA \cong \triangle MBC $ by SAS. Wait, no, SAS would require two sides and included angle. We have one side ($ MD = MB $), one angle ($ \angle MDA = \angle MBC $), and another angle ($ \angle DMA = \angle BMC $). Wait, maybe I made a mistake. Wait, the vertical angles are $ \angle DMA $ and $ \angle BMC $, so in $ \triangle MDA $ and $ \triangle MBC $:

- $ \angle MDA \cong \angle MBC $ (alternate interior)
- $ \overline{MD} \cong \overline{MB} $ (given)
- $ \angle DMA \cong \angle BMC $ (vertical angles)

So that's two angles and a side, but the side is not included? Wait, no, the included side between $ \angle MDA $ and $ \angle DMA $ is $ MD $, and between $ \angle MBC $ and $ \angle BMC $ is $ MB $. Since $ MD = MB $, so that's ASA (angle-side-angle: $ \angle MDA $, $ MD $, $ \angle DMA $ in $ \triangle MDA $; $ \angle MBC $, $ MB $, $ \angle BMC $ in $ \triangle MBC $). So the congruence is $ \triangle MDA \cong \triangle MBC $ by ASA, which is option D. Wait, but let's check the options again:

Option A: $ \triangle MDA \cong \triangle BMC $ by SAS – no, triangles are $ MDA $ and $ MBC $, not $ BMC $.

Option B: $ \triangle MDA \cong \triangle MBC $ by SAS – SAS would need two sides and included angle. We have one side, one angle, one angle. So no.

Option C: $ \triangle MDA \cong \triangle MCB $ by ASA – triangles are $ MDA $ and $ MBC $, not $ MCB $.

Option D: $ \triangle MDA \cong \triangle MBC $ by ASA – yes, because we have two angles ( $ \angle MDA $, $ \angle DMA $) and included side $ MD $ congruent to two angles ( $ \angle MBC $, $ \angle BMC $) and included side $ MB $. So by ASA, $ \triangle MDA \cong \triangle MBC $. Then, by CPCTC, $ AM = CM $. So the missing step is $ \triangle MDA \cong \triangle MBC $ by ASA, which is option D.

Wait, but earlier I thought vertical angles, but actually, the vertical angles are $ \angle DMA $ and $ \angle BMC $, so that's the second angle. So two angles and the included side ( $ MD = MB $ is between the two angles). So ASA. So the correct option is D.
# Answer:
D. $ \triangle MDA \cong \triangle MBC $ by ASA

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QUESTION IMAGE

Question

Explanation:

Step1: Recall SAS and ASA criteria

Step2: Analyze given congruences

Answer:

statements	reasons
\\(\overline{md} \cong \overline{mb}\\)	given
\\(\overline{ad} \parallel \overline{cb}\\)	given
\\(\angle mda \cong \angle mbc\\)	alternate interior angles theorem
?	?
\\(\overline{am} \cong \overline{cm}\\)	cpctc
\\(am = cm\\)	definition of congruence