
3-8 Lengths of the internal and external bisectors
Point 8 of 9 — Unit 3: Parallel Lines, Proportional Segments and Angle BisectorsQuestion bank: 0 questions (0 with pictures)
08 3-8 Lengths of the internal and external bisectors
- Internal: AD² = AB × AC − BD × DC.
- External: AE² = BE × EC − AB × AC (E on the extension of BC).
Check on a 3-4-5 triangle
- Take AB = 3, AC = 4, BC = 5. The internal bisector gives BD = 15/7 and DC = 20/7, so AD² = 12 − 300/49 = 288/49 → AD = 12√2/7.
- The external bisector gives BE = 15, EC = 20, so AE² = 300 − 12 = 288 → AE = 12√2 — note AE > AD, as it must be.
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