Triangle ABC is shown below with line DE passing through points D and E: Triangle ABC is shown with point D lying on side BC and
point E lying on side AC. Line DE is perpendicular to side AC. If triangle ABC is dilated about the center of the triangle to create triangle A'B'C', dilated line D'E' will be parallel to DE be perpendicular to DE lie on the same line as DE shift five units to the right
2 answers:
Answer:
It is given that, A Triangle ABC in which point D lying on side BC and point E lying on side AC.Line DE is perpendicular to side AC.
Now, triangle ABC is dilated about the center of the triangle to create triangle A'B'C' in such a way that D' lies on side B'C' and E' lies on side A'C'.
Both the line segment that is line segment DE and line segment D'E' will be parallel.
Option 1:Dilated line D'E' will be parallel to DE
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Answer:
Transitive property of equality is not a justification for the proof.
Step-by-step explanation:
We draw a right angle ΔACB. CD is perpendicular to AB.
Let AC = a , BC = b , AB = c and CD = h
Now in ΔABC and ΔACD
∠C = ∠D and ∠A = ∠A
from AA similarity postulate
ΔABC similar to ΔACD.
Hence,
=
= c × x ·····················(1)
Now in ΔABC and ΔCBD
∠C = ∠D and ∠B = ∠B
from AA similarity postulates
ΔABC similar to ΔCBD
Hence,
= c × y······················(2)
Add equation (1) and (2)
+
= cx + cy
+
= c(x+y)
+
=
[because x+y=c]
Transitive property is not useful for this proof.
