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 28 Page 423

Congruent triangles have congruent corresponding sides and angles.

A perpendicular bisector, median, and an altitude.

Practice makes perfect

We are told that △ JLM is congruent to △ KLM. Corresponding sides of congruent triangles are congruent. In our case, side JM in △ JLM corresponds to MK in △ KLM. Thus, they are congruent segments and have the same measures. This allows us to conclude that M is a midpoint of JK.

Now, let's review that a median of a triangle is a segment with endpoints being a vertex of a triangle and the midpoint of the opposite side. LM fully satisfies this definition, so it is a median of △ JLK.

We can also use the fact that corresponding angles of congruent triangles are congruent. In △ LMJ, ∠ JML is corresponding to ∠ KML in △ LMK. Hence, these are congruent angles, which means that their measures are the same. m∠ JML=m∠ KML Moreover, these angles are supplementary and the sum of their measures is 180^(∘). m∠ JML+ m∠ KML=180^(∘) Let's substitute m∠ JML for m∠ KML in the above equality and calculate the measure of ∠ JML.
m∠ JML+ m∠ KML=180^(∘)
m∠ JML+ m∠ JML=180^(∘)
2m∠ JML=180^(∘)
m∠ JML=90^(∘)
Therefore, ∠ JML is a right angle and so is ∠ KML. This means that LM is perpendicular to JK. We can conclude that LM is also an altitude of △ JLK.

So far we know that LM is perpendicular to JK and bisects JK into two congruent segments, JM and MK. Therefore, LM is a perpendicular bisector of JK. The final answer is that LM is all three things — a median, an altitude, and a perpendicular bisector of △ JLK.