الأسبوع 10 · منصفا الزاوية والأجزاء المتناسبة: المنصف الداخلي والخارجي وطول المنصف · 3 دروس
2
The bisector of an exterior angle, division inside and outside by the same ratio, and the two bisectors meeting at a right angle
Mathematics Sec 1 Term 1, Unit 3, Lesson 2
GoalUse the exterior-angle bisector, and the fact that the two bisectors at a vertex are perpendicular.
New words
exterior angle · the angle between one side and the extension of another sideexternal division · dividing a segment from a point outside it, in a given ratio
Explanation
The bisector of the exterior angle at A meets the line BC at a point E outside the segment BC, and it divides BC externally in the same ratio: BE/EC = AB/AC. So both bisectors at A divide BC in the ratio AB : AC, one inside (at D) and one outside (at E). The interior and the exterior angles at A together make a straight angle, 180°. Each bisector takes half of its angle, so ∠DAE = ½ × 180° = 90°: the two bisectors are perpendicular. When AB > AC the point E lies beyond C; when AB < AC it lies beyond B. If AB = AC the exterior bisector is parallel to BC and there is no point E. To find EC, remember that BE = BC + EC when E is beyond C, and write the ratio with that.
Worked examples
In △ABC, AB = 12, AC = 8 and BC = 10. The interior bisector AD gives BD/DC = 3/2, so BD = 6, DC = 4. The exterior bisector meets BC at E beyond C: BE/EC = 3/2 with BE = EC + 10, so 2(EC + 10) = 3EC, EC = 20 and BE = 30. DE = DC + CE = 4 + 20 = 24 cm.
Check the right angle. In the same triangle AD² = AB × AC − BD × DC = 96 − 24 = 72 and AE² = BE × EC − AB × AC = 600 − 96 = 504. If ∠DAE = 90°, then DE² = AD² + AE² = 72 + 504 = 576 = 24². This agrees with DE = 24 ✓.
AB = 6, AC = 9 and BC = 5. Now AB < AC, so E is beyond B. EB/EC = 6/9 = 2/3 and EC = EB + 5, so 3EB = 2(EB + 5), EB = 10 and EC = 15.
Class activity
Two folds. On a paper triangle fold the interior bisector at A. Then extend side BA past A with a pencil line and fold the extension onto AC to get the exterior angle bisector. Check with a set square that the two fold lines are perpendicular.
Practice
1
AB = 15, AC = 9, BC = 12. Find BD, DC (interior) and EC, BE, DE (exterior).
Ratio 5/3: BD = 7.5, DC = 4.5. Exterior: BE = EC + 12 and 3(EC + 12) = 5EC, so EC = 18, BE = 30. DE = 4.5 + 18 = 22.5.
2
What happens to the exterior bisector at A when AB = AC?
It is parallel to BC, so there is no point E. (The interior bisector then hits the midpoint of BC.)
3
AB = 4, AC = 6, BC = 5. The exterior bisector meets BC extended at E. Find EB and EC.
AB < AC, so E is beyond B. EB/EC = 2/3 and EC = EB + 5: 3EB = 2EB + 10, EB = 10, EC = 15.
4
Why is ∠DAE = 90°?
The interior angle and the exterior angle at A add up to 180°. The bisectors take half of each: ½ × 180° = 90°.
5
The interior bisector gives BD = 3 and DC = 5. Find the ratio AB : AC, and tell where the exterior bisector meets BC’s line.
AB : AC = 3 : 5. AB < AC, so the exterior bisector meets the line BC beyond B.
