Adding auxiliary lines: Solutions
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1 dding auxiliary lines: Solutions 1. im: To prove b = a + c onstruction: roduce E to meet at F. a E b F c EF = a (corresponding angles, EF) but E = F + F (exterior angle of F) b = a + c 2. im: To prove = 90 onstruction: onstruct O. O O = O = O (radii) O = O (opposite equal sides in ΔO) O = O (opposite equal sides in ΔO) Now in Δ 2 O + 2 O = 180 (angle sum of Δ) = O + O (adjacent angles) = 90 = Education Services ustralia Ltd, except where indicated otherwise. This document may be used, reproduced, published, communicated and adapted free of charge for non-commercial educational purposes provided all acknowledgements associated with the material are retained. page 1 of 6
2 3. im: To prove O = 2. onstruction: onstruct the diameter. O O = O = O (radii) O = O (opposite equal sides in ΔO ) O = O (opposite equal sides in ΔO ) Now in ΔO : O = O + O (exterior angle of ΔO ) = 2 O (1) Similarly in ΔO : O = 2 O (2) ut O = O + O (adjacent angles) = 2 O +2 O (from 1 and 2) = 2 ( O + O ) = 2 (adjacent angles) O = 2 4. ata: = and. im: rove that is a parallelogram. onstruction: onstruct the diagonal. In Δ and Δ 1. is common 2. = (alternate angles, I ) 3. = (data) page 2 of 6
3 Δ Δ (SS) = (matching angles of congruent triangles) ut these are alternate angles! (alternate angles are equal) is a parallelogram (two pairs of opposite sides parallel) 5. ata: = 90 ; = ; bisects im: rove that = +. onstruction: onstruct the perpendicular from to meeting at. In Δ and Δ 1. = ( bisects ) 2. = = 90 (data) 3. is common Δ Δ (S) = (matching sides of congruent triangles) (i) = (matching sides of congruent triangles) (ii) Now in Δ : In Δ : = and = 90 = 45 (equal sides of isosceles Δ) (given) (property of right-angles isosceles triangle) = 180 (angle sum of Δ ) = 45 = (opposite equal angles in Δ ) (iii) Now = + ( collinear) = + (from i) = + (from ii) = + (from iii) = + page 3 of 6
4 hallenge solutions 1. ata: UV im: rove that FG : GH = W : WZ onstruction: onstruct T E F W S U T G Z V H E In ΔSW and ΔTZ 1. SW = TZ (corresponding angles, UV ) 2. WS = ZT (corresponding angles, UV ) ΔSW ΔTZ () S : T = W : Z (matching sides of similar triangles) T : S = Z : W Z = W +WZ (,W, Z collinear points) Similarly, T = S + ST Now T = Z S W S + ST W +WZ becomes = S W ST WZ 1 + = 1 + S W ST WZ = S W i.e. ST : S = WZ : W S : ST = W :WZ SGF is a parallelogram (2 pairs of opposite sides parallel) S = FG (opposite sides of parallelogram) Similarly, STHG is a parallelogram and GH = ST. FG :GH = S : ST = W :WZ Note: This result is the theorem: arallel lines preserve ratios of intercepts on transversals. It can be abbreviated as parallel lines preserve ratios. page 4 of 6
5 2. ata: bisects im: rove that =. onstruction: roduce to so that. From uestion 1, as then : = : Now = and = (corresponding angles, ) (alternate angles, ) ut = ( bisects ) = Δ is isosceles = (opposite equal angles in Δ ) = (parallel lines preserve ratios) i.e. the result from u. 1. ut = (proven above) = Note: This result is a corollary of the theorem proven in uestion 1. n alternative proof, which does not use the theorem, appears on the following page. page 5 of 6
6 lternatively: ata: bisects im: rove that =. onstruction: roduce to so that. Let = x Then = x = + = 2 = 2x = x = ( bisects ) (opposite equal sides in isosceles Δ) (exterior angle of Δ) In Δ and Δ 1. is common 2. = (proven above) Δ Δ () = = (matching sides of similar triangles) Taking = : = ( + ) = ( + ) ( and are collinear) + = + = ut = by construction = = (dividing throughout by ) page 6 of 6
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