Similarity proofs: fill in the missing step or reason
Grade 10 Math Worksheet

Circle the best answer.
Read this first
Use this
Given: DH ∥ FG. Prove: DQ/GQ = HQ/FQ. Ladder: 1. DH ∥ FG -- Given; 2. ∠HDQ ≅ ∠FGQ -- Alternate interior angles theorem; 3. ∠DQH ≅ ∠GQF -- Vertical angles theorem; 4. triangle DQH ~ triangle GQF -- AA similarity postulate; 5. DQ/GQ = HQ/FQ -- ?.
Bo draws a bike frame where bars DG and FH cross at Q, with DH ∥ FG. Which reason fills the blank?
- Definition of similar triangles
- AA similarity postulate
- Triangle proportionality theorem
- Given
Esme crosses craft sticks SV and UW at C, with SW ∥ UV. Which reason fills the blank?
Sketch, not to scale: segments SV and UW cross at C (S upper left, V lower right, W upper right, U lower left). Marked: arrows on SW and UV. Given: SW ∥ UV. Prove: SC/VC = WC/UC. Ladder: 1. SW ∥ UV — Given; 2. ∠WSC ≅ ∠UVC — Alternate interior angles theorem; 3. ∠SCW ≅ ∠VCU — Vertical angles theorem; 4. △SCW ~ △VCU — AA similarity postulate; 5. SC/VC = WC/UC — ?. Sketch, not to scale: segments SV and UW cross at C (S upper left, V lower right, W upper right, U lower left). Marked: arrows on SW and UV. Given: SW ∥ UV. Prove: SC/VC = WC/UC. Ladder: 1. SW ∥ UV — Given; 2. ∠WSC ≅ ∠UVC — Alternate interior angles theorem; 3. ∠SCW ≅ ∠VCU — Vertical angles theorem; 4. △SCW ~ △VCU — AA similarity postulate; 5. SC/VC = WC/UC — ?. S W C V U Given: SW ∥ UV. 1. SW ∥ UV -- Given; 2. ∠WSC ≅ ∠UVC -- Alternate interior angles theorem; 3. ∠SCW ≅ ∠VCU -- Vertical angles theorem; 4. triangle SCW ~ triangle VCU -- AA similarity postulate; 5. SC/VC = WC/UC -- ?. Not to scale. - Definition of similar triangles
- Given
- Triangle proportionality theorem
- AA similarity postulate
Use this
Given: HD ∥ YU. Prove: HF/UF = DF/YF. Ladder: 1. HD ∥ YU -- Given; 2. ∠DHF ≅ ∠YUF -- Alternate interior angles theorem; 3. ∠HFD ≅ ∠UFY -- Vertical angles theorem; 4. triangle HFD ~ triangle UFY -- ?; 5. HF/UF = DF/YF -- Definition of similar triangles.
Jun opens scissors with blades HU and YD crossing at F, with HD ∥ YU. Which reason fills the blank?
Sketch, not to scale: segments HU and YD cross at F (H upper left, U lower right, D upper right, Y lower left). Marked: arrows on HD and YU. Given: HD ∥ YU. Prove: HF/UF = DF/YF. Ladder: 1. HD ∥ YU — Given; 2. ∠DHF ≅ ∠YUF — Alternate interior angles theorem; 3. ∠HFD ≅ ∠UFY — Vertical angles theorem; 4. △HFD ~ △UFY — ?; 5. HF/UF = DF/YF — Definition of similar triangles. Sketch, not to scale: segments HU and YD cross at F (H upper left, U lower right, D upper right, Y lower left). Marked: arrows on HD and YU. Given: HD ∥ YU. Prove: HF/UF = DF/YF. Ladder: 1. HD ∥ YU — Given; 2. ∠DHF ≅ ∠YUF — Alternate interior angles theorem; 3. ∠HFD ≅ ∠UFY — Vertical angles theorem; 4. △HFD ~ △UFY — ?; 5. HF/UF = DF/YF — Definition of similar triangles. HD ∥ YU H U D Y F Not to scale. - AAS
- Vertical angles theorem
- SAS similarity theorem
- AA similarity postulate
Use this
Ladder: 1. WC ∥ GD -- Given; 2. ∠CWS ≅ ∠GDS -- Alternate interior angles theorem; 3. ∠WSC ≅ ∠DSG -- Vertical angles theorem; 4. triangle WSC ~ triangle DSG -- ?; 5. WS/DS = CS/GS -- Definition of similar triangles.
Kira draws two ramps WD and GC that cross at S, with WC ∥ GD. Which reason fills the blank?
Sketch, not to scale: segments WD and GC cross at S (W upper left, D lower right, C upper right, G lower left). Marked: arrows on WC and GD. Given: WC ∥ GD. Prove: WS/DS = CS/GS. Ladder: 1. WC ∥ GD — Given; 2. ∠CWS ≅ ∠GDS — Alternate interior angles theorem; 3. ∠WSC ≅ ∠DSG — Vertical angles theorem; 4. △WSC ~ △DSG — ?; 5. WS/DS = CS/GS — Definition of similar triangles. Sketch, not to scale: segments WD and GC cross at S (W upper left, D lower right, C upper right, G lower left). Marked: arrows on WC and GD. Given: WC ∥ GD. Prove: WS/DS = CS/GS. Ladder: 1. WC ∥ GD — Given; 2. ∠CWS ≅ ∠GDS — Alternate interior angles theorem; 3. ∠WSC ≅ ∠DSG — Vertical angles theorem; 4. △WSC ~ △DSG — ?; 5. WS/DS = CS/GS — Definition of similar triangles. W C S D G Given: WC ∥ GD. Prove: WS/DS = CS/GS. Not to scale. - ASA
- Vertical angles theorem
- AA similarity postulate
- SSS similarity theorem
Luz measures a stripe VQ across a road sign HCS, level with its base CS. Which reason fills the blank?
Sketch, not to scale: triangle HCS with H at the top and base CS; V is on HC, Q is on HS, and segment VQ is drawn. Marked: arrows on VQ and CS; HV = 4, VC = 3, HQ = 8. Given: VQ ∥ CS, HV = 4, VC = 3 and HQ = 8. Prove: QS = 6. Ladder: 1. VQ ∥ CS — Given; 2. HV/VC = HQ/QS — Triangle proportionality theorem; 3. 4/3 = 8/QS — Substitution property; 4. 4 · QS = 24 — ?; 5. QS = 6 — Division property of equality. Sketch, not to scale: triangle HCS with H at the top and base CS; V is on HC, Q is on HS, and segment VQ is drawn. Marked: arrows on VQ and CS; HV = 4, VC = 3, HQ = 8. Given: VQ ∥ CS, HV = 4, VC = 3 and HQ = 8. Prove: QS = 6. Ladder: 1. VQ ∥ CS — Given; 2. HV/VC = HQ/QS — Triangle proportionality theorem; 3. 4/3 = 8/QS — Substitution property; 4. 4 · QS = 24 — ?; 5. QS = 6 — Division property of equality. C S H V Q 4 3 8 Not to scale - Division property of equality
- Triangle proportionality theorem
- Substitution property
- Cross products property
Use this
Given: YF ∥ VC. Prove: YH/CH = FH/VH. Ladder: 1. YF ∥ VC -- Given; 2. ? -- Alternate interior angles theorem; 3. ∠YHF ≅ ∠CHV -- Vertical angles theorem; 4. triangle YHF ~ triangle CHV -- AA similarity postulate; 5. YH/CH = FH/VH -- Definition of similar triangles.
Noah draws scissors whose blades YC and VF cross at H, with YF ∥ VC. Which statement fills the blank?
Sketch, not to scale: segments YC and VF cross at H (Y upper left, C lower right, F upper right, V lower left). Marked: arrows on YF and VC. Given: YF ∥ VC. Prove: YH/CH = FH/VH. Ladder: 1. YF ∥ VC — Given; 2. ? — Alternate interior angles theorem; 3. ∠YHF ≅ ∠CHV — Vertical angles theorem; 4. △YHF ~ △CHV — AA similarity postulate; 5. YH/CH = FH/VH — Definition of similar triangles. Sketch, not to scale: segments YC and VF cross at H (Y upper left, C lower right, F upper right, V lower left). Marked: arrows on YF and VC. Given: YF ∥ VC. Prove: YH/CH = FH/VH. Ladder: 1. YF ∥ VC — Given; 2. ? — Alternate interior angles theorem; 3. ∠YHF ≅ ∠CHV — Vertical angles theorem; 4. △YHF ~ △CHV — AA similarity postulate; 5. YH/CH = FH/VH — Definition of similar triangles. Y F C V H Not to scale. - ∠FYH ≅ ∠CVH
- ∠FYH ≅ ∠VCH
- ∠FYH ≅ ∠YFH
- YH/CH = FH/VH
Use this
Sketch, not to scale: segments UH and SQ cross at G (U upper left, H lower right, Q upper right, S lower left). Marked: arrows on UQ and SH. Given: UQ ∥ SH. Prove: UG/HG = QG/SG. Ladder: 1. UQ ∥ SH -- Given; 2. ∠QUG ≅ ∠SHG -- Alternate interior angles theorem; 3. ∠UGQ ≅ ∠HGS -- Vertical angles theorem; 4. triangle UGQ ~ triangle HGS -- AA similarity postulate; 5. ? -- Definition of similar triangles.
Pia draws bike-rack bars UH and SQ that cross at G, with UQ ∥ SH. Which statement fills the blank?
Sketch, not to scale: segments UH and SQ cross at G (U upper left, H lower right, Q upper right, S lower left). Marked: arrows on UQ and SH. Given: UQ ∥ SH. Prove: UG/HG = QG/SG. Ladder: 1. UQ ∥ SH — Given; 2. ∠QUG ≅ ∠SHG — Alternate interior angles theorem; 3. ∠UGQ ≅ ∠HGS — Vertical angles theorem; 4. △UGQ ~ △HGS — AA similarity postulate; 5. ? — Definition of similar triangles. Sketch, not to scale: segments UH and SQ cross at G (U upper left, H lower right, Q upper right, S lower left). Marked: arrows on UQ and SH. Given: UQ ∥ SH. Prove: UG/HG = QG/SG. Ladder: 1. UQ ∥ SH — Given; 2. ∠QUG ≅ ∠SHG — Alternate interior angles theorem; 3. ∠UGQ ≅ ∠HGS — Vertical angles theorem; 4. △UGQ ~ △HGS — AA similarity postulate; 5. ? — Definition of similar triangles. U Q S H G Not to scale. - UQ/SH = UG/SG
- UG/HG = SG/QG
- UG ≅ HG
- UG/HG = QG/SG
Ren measures a cut HU across a sandwich half WQC, level with the edge QC. Which statement fills the blank?
Sketch, not to scale: triangle WQC with W at the top and base QC; H is on WQ, U is on WC, and segment HU is drawn. Marked: arrows on HU and QC; WH = 3, HQ = 5, WU = 9. Given: HU ∥ QC, WH = 3, HQ = 5 and WU = 9. Prove: UC = 15. Ladder: 1. HU ∥ QC — Given; 2. WH/HQ = WU/UC — Triangle proportionality theorem; 3. ? — Substitution property; 4. 3 · UC = 45 — Cross products property; 5. UC = 15 — Division property of equality. Sketch, not to scale: triangle WQC with W at the top and base QC; H is on WQ, U is on WC, and segment HU is drawn. Marked: arrows on HU and QC; WH = 3, HQ = 5, WU = 9. Given: HU ∥ QC, WH = 3, HQ = 5 and WU = 9. Prove: UC = 15. Ladder: 1. HU ∥ QC — Given; 2. WH/HQ = WU/UC — Triangle proportionality theorem; 3. ? — Substitution property; 4. 3 · UC = 45 — Cross products property; 5. UC = 15 — Division property of equality. Q C W H U 3 5 9 Not to scale - 3/5 = 9/UC
- UC = 15
- 3/8 = 9/UC
- 3/5 = UC/9
Use this
Given: CY ∥ QW. Prove: CV/WV = YV/QV. Ladder: 1. CY ∥ QW -- Given; 2. ∠YCV ≅ ∠QWV -- Alternate interior angles theorem; 3. ∠CVY ≅ ∠WVQ -- Vertical angles theorem; 4. triangle CVY ~ triangle WVQ -- AA similarity postulate; 5. ? -- Definition of similar triangles.
Tala crosses craft sticks CW and QY at V, with CY ∥ QW. Which statement fills the blank?
Sketch, not to scale: segments CW and QY cross at V (C upper left, W lower right, Y upper right, Q lower left). Marked: arrows on CY and QW. Given: CY ∥ QW. Prove: CV/WV = YV/QV. Ladder: 1. CY ∥ QW — Given; 2. ∠YCV ≅ ∠QWV — Alternate interior angles theorem; 3. ∠CVY ≅ ∠WVQ — Vertical angles theorem; 4. △CVY ~ △WVQ — AA similarity postulate; 5. ? — Definition of similar triangles. Sketch, not to scale: segments CW and QY cross at V (C upper left, W lower right, Y upper right, Q lower left). Marked: arrows on CY and QW. Given: CY ∥ QW. Prove: CV/WV = YV/QV. Ladder: 1. CY ∥ QW — Given; 2. ∠YCV ≅ ∠QWV — Alternate interior angles theorem; 3. ∠CVY ≅ ∠WVQ — Vertical angles theorem; 4. △CVY ~ △WVQ — AA similarity postulate; 5. ? — Definition of similar triangles. C Y Q W V Not to scale. - CV/WV = YV/QV
- CY/QW = CV/QV
- CV/WV = QV/YV
- CV ≅ WV
Zee draws scissors whose blades FS and HG cross at D, with FG ∥ HS. Which statement fills the blank?
Sketch, not to scale: segments FS and HG cross at D (F upper left, S lower right, G upper right, H lower left). Marked: arrows on FG and HS. Given: FG ∥ HS. Prove: FD/SD = GD/HD. Ladder: 1. FG ∥ HS — Given; 2. ∠GFD ≅ ∠HSD — Alternate interior angles theorem; 3. ∠FDG ≅ ∠SDH — Vertical angles theorem; 4. ? — AA similarity postulate; 5. FD/SD = GD/HD — Definition of similar triangles. Sketch, not to scale: segments FS and HG cross at D (F upper left, S lower right, G upper right, H lower left). Marked: arrows on FG and HS. Given: FG ∥ HS. Prove: FD/SD = GD/HD. Ladder: 1. FG ∥ HS — Given; 2. ∠GFD ≅ ∠HSD — Alternate interior angles theorem; 3. ∠FDG ≅ ∠SDH — Vertical angles theorem; 4. ? — AA similarity postulate; 5. FD/SD = GD/HD — Definition of similar triangles. Sketch, not to scale F S G H D Statement Reason 1. FG ∥ HS Given 2. ∠GFD ≅ ∠HSD Alternate interior angles theorem 3. ∠FDG ≅ ∠SDH Vertical angles theorem 4. ? AA similarity postulate 5. FD/SD = GD/HD Definition of similar triangles - △FDG ~ △FDH
- △FDG ~ △SDH
- △FDG ≅ △SDH
- △FDG ~ △HDS