Answer:
- reflection across BC
- the image of a vertex will coincide with its corresponding vertex
- SSS: AB≅GB, AC≅GC, BC≅BC.
Step-by-step explanation:
We want to identify a rigid transformation that maps congruent triangles to one-another, to explain the coincidence of corresponding parts, and to identify the theorems that show congruence.
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<h3>1.</h3>
Triangles GBC and ABC share side BC. Whatever rigid transformation we use will leave segment BC invariant. Translation and rotation do not do that. The only possible transformation that will leave BC invariant is <em>reflection across line BC</em>.
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<h3>2.</h3>
In part 3, we show ∆GBC ≅ ∆ABC. That means vertices A and G are corresponding vertices. When we map the congruent figures onto each other, <em>corresponding parts are coincident</em>. That is, vertex G' (the image of vertex G) will coincide with vertex A.
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<h3>3.</h3>
The markings on the figure show the corresponding parts to be ...
- side AB and side GB
- side AC and side GC
- angle ABC and angle GBC
- angle BAC and angle BGC
And the reflexive property of congruence tells us BC corresponds to itself:
There are four available congruence theorems applicable to triangles that are not right triangles
- SSS -- three pairs of corresponding sides
- SAS -- two corresponding sides and the angle between
- ASA -- two corresponding angles and the side between
- AAS -- two corresponding angles and the side not between
We don't know which of these are in your notes, but we do know that all of them can be used. AAS can be used with two different sides. SAS can be used with two different angles.
SSS
Corresponding sides are listed above. Here, we list them again:
AB and GB; AC and GC; BC and BC
SAS
One use is with AB, BC, and angle ABC corresponding to GB, BC, and angle GBC.
Another use is with BA, AC, and angle BAC corresponding to BG, GC, and angle BGC.
ASA
Angles CAB and CBA, side AB corresponding to angles CGB and CBG, side GB.
AAS
One use is with angles CBA and CAB, side CB corresponding to angles CBG and CGB, side CB.
Another use is with angles CBA and CAB, side CA corresponding to angles CBG and CGB, side CG.
, this equation is slope intercept for, which is
so our m or slope is
then we use the equation
, so what do we multiply
by to get -1? -4 right, so that's the slope of line AB with the point (2,0)
so m= -4 and our point is (-1,6)
thats the equation of line side CD
