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Since EC is a median of △ AED, segments AC and CD are congruent. EC is an altitude if ∠ACE is a right angle.
x=6, EC is not an altitude. See solution.
Let's solve this exercise by first calculating the value of x and then determining if EC is an altitude of â–³ AED.
It is given that EC is a median of â–³ AED. Let's recall that a median of a triangle is a segment with endpoints being a vertex of a triangle and the midpoint of the opposite side. Thus, C is a midpoint of AD. By the definition of a midpoint, segments AC and CD are congruent.
AC= 4x-3, DC= 2x+9
LHS-2x=RHS-2x
LHS+3=RHS+3
.LHS /2.=.RHS /2.
An altitude of a triangle is a segment from a vertex to the line containing the opposite side and perpendicular to the line containing that side. Thus, if EC is perpendicular to AD, then we can conclude it is an altitude of △ AED. Remember, perpendicular lines form right angles. Let's check if angle ∠ECA is a right angle and measures 90^(∘).
We are given that ∠ECA measures 15x+2. Let's substitute x with 6 into this expression and calculate m∠ECA.
x= 6
Multiply
Add terms
As we can see, ∠ECA measures 92^(∘) and not 90^(∘). It's not a right angle. Therefore, EC is not an altitude of △ AED.