Answer: if Gregory draws the segment with endpoints A and A’, then the midpoint will lie on the line of reflection.
Explanation:
Given that a triangle ABC is reflected in triangle A'B'C'
Here reflection is done on a line
If you imagine the line as a mirror then ABC will have image on the mirror line as A'B'C'
Recall that in a mirror the object and image would be equidistant from the mirror and also the line joining the image and object would be perpendicular to the mirror
But note that corresponding images will only be perpendicular bisector to the line
So A and A' only will be corresponding so AA' will have mid point on line
Option 1 is right
Your LCM is 11880 , and your HCF is 36
Answer:
C.
Step-by-step explanation:
If Nick starts with 20, we can just plug it into p/2 - 12 to get 10-12 or -2. This means Nick did not get 20 cards. Now we have answered both. Two-liner!
I draw the two triangles, see the picture attached.
As you can see, angle 1 and 2 are vertically opposite angles because they are formed by the same two crossing lines and they face each other.
Angles <span>ABQ and QPR, as well as angles BAQ and QRP, are alternate interior angles because they are formed by </span><span>two parallel lines crossed by a transversal, and they are inside the two lines on opposite sides of the transversal.</span>
Hence, Allison's correct claims are:
1 = 2 because they are vertically opposite angles. BAQ = QRP because they are alternate interior angles. Therefore Allison, in order to prove her claim, can use the AA similarity theorem: if two angles of a triangle are congruent to two angles of the other triangle, then the two triangles are similar.