Special Segments in Triangles

Medians, altitudes, and angle bisectors

Special segments in triangles content

📚 Practice Problems

1Problem 1easy

❓ Question:

In triangle ABC, point M is the midpoint of side BC. If BC = 18, find BM.

💡 Show Solution

Step 1: Understand what a midpoint is: A midpoint divides a segment into two equal parts

Step 2: Since M is the midpoint of BC: BM = MC BM + MC = BC

Step 3: Substitute: BM + BM = 18 2(BM) = 18

Step 4: Solve: BM = 18/2 BM = 9

Step 5: Verify: BM = 9 and MC = 9 BM + MC = 9 + 9 = 18 ✓

Answer: BM = 9

2Problem 2easy

❓ Question:

A median of a triangle connects a vertex to the midpoint of the opposite side. In triangle DEF, the median from D to side EF has length 12. If the centroid is at point G, find DG.

💡 Show Solution

Step 1: Recall the centroid property: The centroid divides each median in a 2:1 ratio The distance from vertex to centroid is 2/3 of the median The distance from centroid to midpoint is 1/3 of the median

Step 2: Identify given information: Median from D = 12 Centroid = G

Step 3: Find DG (vertex to centroid): DG = (2/3) × median DG = (2/3) × 12 DG = 24/3 DG = 8

Step 4: Find the other part (for verification): Distance from G to midpoint = (1/3) × 12 = 4

Step 5: Verify: 8 + 4 = 12 ✓ 8/4 = 2/1 ✓ (2:1 ratio)

Answer: DG = 8

3Problem 3medium

❓ Question:

In triangle ABC, the perpendicular bisector of side AB passes through point P. If PA = 15, find PB.

💡 Show Solution

Step 1: Recall perpendicular bisector property: A perpendicular bisector of a segment is perpendicular to the segment and passes through its midpoint

Step 2: Key property of perpendicular bisectors: Any point on the perpendicular bisector is equidistant from the endpoints of the segment

Step 3: Apply the property: Since P is on the perpendicular bisector of AB: PA = PB

Step 4: Solve: PA = 15 (given) Therefore: PB = 15

Step 5: Conceptual understanding: This makes sense because the perpendicular bisector is the set of all points equidistant from A and B

Answer: PB = 15

4Problem 4medium

❓ Question:

In triangle XYZ, an altitude is drawn from X to side YZ, meeting it at point H. If angle XHY = 90°, XH = 12, and YH = 5, find XY.

💡 Show Solution

Step 1: Understand the altitude: An altitude is perpendicular to the side it meets So angle XHY = 90° confirms XH is an altitude

Step 2: Identify the right triangle: Triangle XHY is a right triangle with:

  • Right angle at H
  • One leg XH = 12
  • Other leg YH = 5
  • Hypotenuse XY = ?

Step 3: Use Pythagorean Theorem: XH² + YH² = XY² 12² + 5² = XY² 144 + 25 = XY² 169 = XY²

Step 4: Solve for XY: XY = √169 XY = 13

Step 5: Recognize the Pythagorean triple: This is a 5-12-13 right triangle

Answer: XY = 13

5Problem 5hard

❓ Question:

In triangle ABC, the three medians intersect at centroid G. The median from vertex A has length 18, the median from vertex B has length 15, and the median from vertex C has length 21. Find the distance from each vertex to the centroid.

💡 Show Solution

Step 1: Recall centroid property: The centroid divides each median in a 2:1 ratio Distance from vertex to centroid = (2/3) × median length

Step 2: Find distance from A to G: AG = (2/3) × 18 AG = 36/3 AG = 12

Step 3: Find distance from B to G: BG = (2/3) × 15 BG = 30/3 BG = 10

Step 4: Find distance from C to G: CG = (2/3) × 21 CG = 42/3 CG = 14

Step 5: Verify the 2:1 ratio for each: Median from A: vertex to centroid = 12, centroid to midpoint = 6 12:6 = 2:1 ✓ Median from B: vertex to centroid = 10, centroid to midpoint = 5 10:5 = 2:1 ✓ Median from C: vertex to centroid = 14, centroid to midpoint = 7 14:7 = 2:1 ✓

Step 6: Summary: The centroid is the "balance point" of the triangle It's located 2/3 of the way from each vertex to the opposite side's midpoint

Answer: AG = 12, BG = 10, CG = 14

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