In ∆ABC, the altitudes from vertex B and C intersect at point M, so that BM = CM. Prove that ∆ABC is isosceles.
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Hey there!
To solve this, we can use division. First, lets turn into an improper fraction:
So, we have
We can multiply by the reciprocal:
That gives us
Hope it helps and have a great day!
A sequence of transformations that maps △DEF to △D′E′F′ is a rotation of 90° counterclockwise about the origin followed by a translation two units right.
<h3>What is the sequence of transformations?</h3>The sequence of vertices ABC(DEF in this question) is clockwise, as is the sequence of A'B'C'(D'E'F in this question). Thus, an even number of reflections is involved, if any reflections are involved. The offered choices do not include suitable reflections.
The orientation of AB(DE) is toward the right. The orientation of A'B'(D'E') is up, so there must be a rotation of 90° CCW. Rotation of 90° CCW about the origin will leave the figure in a position that is 2 units left of where it is shown. The rotation must be followed by a translation 2 units to the right.
Thus, we conclude that a sequence of transformations that maps △DEF to △D′E′F′ is a rotation of 90° counterclockwise about the origin followed by a translation two units right.
Read more about Sequence of Transformations at; brainly.com/question/4289712
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