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2 answers:
Question 1 is correct. AM = 4 and CM cong. AM therefore CM = 4.
Question 2 is correct. Triangles congruent by SAS --> BA cong BC = 12
Question 3 Line BM intersects point D
Questions 4-6 Attachement 1 is an example with these triangles and their circumcenters as det. by their perp. bisectors. As you can see:
The POC of a right triangle lies on the triangle, acute, inside, and obtuse, outside.
Question 7-8 First off: DF = 6, not BD.
We can clearly see 6 congruent triangles within this equilateral one.
BD corresponds to AD and = 11.5
So does DC.
Question 9-10 See attachement 2
Question 11 Sometimes -- see the right triangle in attachment 1
Question 12 Never -- See attachment 4, of course you can always just draw one for yourself. Just draw an acute, right, and obtuse triangle!
Question 13 Always -- See attahcment 3
Question 14 Never -- See attachment 3
Question 15 Sometimes -- see obtuse triangle in attachment 1
Question 2 is correct. Triangles congruent by SAS --> BA cong BC = 12
Question 3 Line BM intersects point D
Questions 4-6 Attachement 1 is an example with these triangles and their circumcenters as det. by their perp. bisectors. As you can see:
The POC of a right triangle lies on the triangle, acute, inside, and obtuse, outside.
Question 7-8 First off: DF = 6, not BD.
We can clearly see 6 congruent triangles within this equilateral one.
BD corresponds to AD and = 11.5
So does DC.
Question 9-10 See attachement 2
Question 11 Sometimes -- see the right triangle in attachment 1
Question 12 Never -- See attachment 4, of course you can always just draw one for yourself. Just draw an acute, right, and obtuse triangle!
Question 13 Always -- See attahcment 3
Question 14 Never -- See attachment 3
Question 15 Sometimes -- see obtuse triangle in attachment 1
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