McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
2. Medians and Altitudes of Triangles
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Exercise 25 Page 423

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.

Practice makes perfect

Let's solve this exercise by first calculating the value of x and then determining if EC is an altitude of △ AED.

Calculate x

It is given that EC is a median of △ 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.

Congruent segments have the same measures. Therefore, the following equality is true. AC=CD We are given that AC measures 4x-3 and DC measures 2x+9. If we substitute these expressions into the above equality, we will get the equation for x. Let's solve it!
AC=CD
4x-3= 2x+9
2x-3=9
2x=12
x=6

Is EC an Altitude of △ AED?

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.
m∠ ECA=15x+2
m∠ ECA=15( 6)+2
m∠ ECA=90+2
m∠ ECA=92
As we can see, ∠ ECA measures 92^(∘) and not 90^(∘). It's not a right angle. Therefore, EC is not an altitude of △ AED.