Triangle proofs: fill in the missing step or reason
Grade 10 Math Worksheet - Challenge Me

Circle the best answer.
Kira sews a seam SY on a sail SUW; the two bottom pieces have lengths written with x. Which reason fills the blank?
- Multiplication property of equality
- Addition property of equality
- Definition of midpoint
- Subtraction property of equality
Luz hangs a string HQ inside a tent front HGV; the two floor pieces have lengths written with x. Which reason fills the blank?
Sketch, not to scale: triangle HGV with H at the top and base GV; segment HQ runs from H to point Q on GV. Marked: QV labeled 6x − 4; GQ labeled 3x + 11. Given: HQ is a median of △HGV, QV = 6x − 4 and GQ = 3x + 11. Prove: x = 5. Ladder: 1. HQ is a median of △HGV — Given; 2. Q is the midpoint of GV — Definition of median; 3. QV = GQ — Definition of midpoint; 4. 6x − 4 = 3x + 11 — Substitution property; 5. 3x − 4 = 11 — Subtraction property of equality; 6. 3x = 15 — Addition property of equality; 7. x = 5 — ?. Sketch, not to scale: triangle HGV with H at the top and base GV; segment HQ runs from H to point Q on GV. Marked: QV labeled 6x − 4; GQ labeled 3x + 11. Given: HQ is a median of △HGV, QV = 6x − 4 and GQ = 3x + 11. Prove: x = 5. Ladder: 1. HQ is a median of △HGV — Given; 2. Q is the midpoint of GV — Definition of median; 3. QV = GQ — Definition of midpoint; 4. 6x − 4 = 3x + 11 — Substitution property; 5. 3x − 4 = 11 — Subtraction property of equality; 6. 3x = 15 — Addition property of equality; 7. x = 5 — ?. G V H Q 6x − 4 3x + 11 Not to scale - Subtraction property of equality
- Substitution property
- Given
- Division property of equality
Milo cuts a triangle sandwich DCY along median DS; the two edge pieces have lengths written with x. Which reason fills the blank?
Sketch, not to scale: triangle DCY with D at the top and base CY; segment DS runs from D to point S on CY. Marked: SY labeled 7x − 12; CS labeled 3x + 8. Given: DS is a median of △DCY, SY = 7x − 12 and CS = 3x + 8. Prove: x = 5. Ladder: 1. DS is a median of △DCY — Given; 2. S is the midpoint of CY — Definition of median; 3. SY = CS — Definition of midpoint; 4. 7x − 12 = 3x + 8 — Substitution property; 5. 4x − 12 = 8 — Subtraction property of equality; 6. 4x = 20 — Addition property of equality; 7. x = 5 — ?. Sketch, not to scale: triangle DCY with D at the top and base CY; segment DS runs from D to point S on CY. Marked: SY labeled 7x − 12; CS labeled 3x + 8. Given: DS is a median of △DCY, SY = 7x − 12 and CS = 3x + 8. Prove: x = 5. Ladder: 1. DS is a median of △DCY — Given; 2. S is the midpoint of CY — Definition of median; 3. SY = CS — Definition of midpoint; 4. 7x − 12 = 3x + 8 — Substitution property; 5. 4x − 12 = 8 — Subtraction property of equality; 6. 4x = 20 — Addition property of equality; 7. x = 5 — ?. C Y D S 7x − 12 3x + 8 Not to scale - Division property of equality
- Given
- Addition property of equality
- Definition of median
Read this first
Use this
Given: CD ≅ CF, and CQ is a median of triangle CDF. Prove: CQ bisects ∠DCF. Ladder: 1. CD ≅ CF -- Given; 2. CQ is a median of triangle CDF -- Given; 3. Q is the midpoint of DF -- Definition of median; 4. ? -- Definition of midpoint; 5. CQ ≅ CQ -- Reflexive property; 6. triangle CDQ ≅ triangle CFQ -- SSS; 7. ∠DCQ ≅ ∠FCQ -- CPCTC; 8. CQ bisects ∠DCF -- Definition of angle bisector.
Noah shows that the median CQ of road sign CDF splits its top angle evenly. Which statement fills the blank?
Sketch, not to scale: triangle CDF with C at the top and base DF; segment CQ runs from C to point Q on DF. Marked: CD and CF with one tick each. Given: CD ≅ CF, and CQ is a median of △CDF. Prove: CQ bisects ∠DCF. Ladder: 1. CD ≅ CF — Given; 2. CQ is a median of △CDF — Given; 3. Q is the midpoint of DF — Definition of median; 4. ? — Definition of midpoint; 5. CQ ≅ CQ — Reflexive property; 6. △CDQ ≅ △CFQ — SSS; 7. ∠DCQ ≅ ∠FCQ — CPCTC; 8. CQ bisects ∠DCF — Definition of angle bisector. Sketch, not to scale: triangle CDF with C at the top and base DF; segment CQ runs from C to point Q on DF. Marked: CD and CF with one tick each. Given: CD ≅ CF, and CQ is a median of △CDF. Prove: CQ bisects ∠DCF. Ladder: 1. CD ≅ CF — Given; 2. CQ is a median of △CDF — Given; 3. Q is the midpoint of DF — Definition of median; 4. ? — Definition of midpoint; 5. CQ ≅ CQ — Reflexive property; 6. △CDQ ≅ △CFQ — SSS; 7. ∠DCQ ≅ ∠FCQ — CPCTC; 8. CQ bisects ∠DCF — Definition of angle bisector. C D F Q Not to scale. - DQ ≅ FQ
- CD ≅ DQ
- CQ ≅ DF
- CQ bisects ∠DCF
Use this
Sketch, not to scale: triangle SWY with S at the top and base WY; segment SU runs from S to point U on WY. Marked: SW and SY with one tick each. Given: SW ≅ SY, and SU is a median of triangle SWY. Prove: SU bisects ∠WSY. Ladder: 1. SW ≅ SY -- Given; 2. SU is a median of triangle SWY -- Given; 3. U is the midpoint of WY -- Definition of median; 4. ? -- Definition of midpoint; 5. SU ≅ SU -- Reflexive property; 6. triangle SWU ≅ triangle SYU -- SSS; 7. ∠WSU ≅ ∠YSU -- CPCTC; 8. SU bisects ∠WSY -- Definition of angle bisector.
Pia shows that the median SU of roof SWY splits its top angle evenly. Which statement fills the blank?
Sketch, not to scale: triangle SWY with S at the top and base WY; segment SU runs from S to point U on WY. Marked: SW and SY with one tick each. Given: SW ≅ SY, and SU is a median of △SWY. Prove: SU bisects ∠WSY. Ladder: 1. SW ≅ SY — Given; 2. SU is a median of △SWY — Given; 3. U is the midpoint of WY — Definition of median; 4. ? — Definition of midpoint; 5. SU ≅ SU — Reflexive property; 6. △SWU ≅ △SYU — SSS; 7. ∠WSU ≅ ∠YSU — CPCTC; 8. SU bisects ∠WSY — Definition of angle bisector. Sketch, not to scale: triangle SWY with S at the top and base WY; segment SU runs from S to point U on WY. Marked: SW and SY with one tick each. Given: SW ≅ SY, and SU is a median of △SWY. Prove: SU bisects ∠WSY. Ladder: 1. SW ≅ SY — Given; 2. SU is a median of △SWY — Given; 3. U is the midpoint of WY — Definition of median; 4. ? — Definition of midpoint; 5. SU ≅ SU — Reflexive property; 6. △SWU ≅ △SYU — SSS; 7. ∠WSU ≅ ∠YSU — CPCTC; 8. SU bisects ∠WSY — Definition of angle bisector. S W Y U Not to scale. - SW ≅ WU
- SU ≅ WY
- SU bisects ∠WSY
- WU ≅ YU
Ren shows that the median GD of tent front GHV splits its top angle evenly. Which statement fills the blank?
Sketch, not to scale: triangle GHV with G at the top and base HV; segment GD runs from G to point D on HV. Marked: GH and GV with one tick each. Given: GH ≅ GV, and GD is a median of △GHV. Prove: GD bisects ∠HGV. Ladder: 1. GH ≅ GV — Given; 2. GD is a median of △GHV — Given; 3. D is the midpoint of HV — Definition of median; 4. HD ≅ VD — Definition of midpoint; 5. GD ≅ GD — Reflexive property; 6. ? — SSS; 7. ∠HGD ≅ ∠VGD — CPCTC; 8. GD bisects ∠HGV — Definition of angle bisector. Sketch, not to scale: triangle GHV with G at the top and base HV; segment GD runs from G to point D on HV. Marked: GH and GV with one tick each. Given: GH ≅ GV, and GD is a median of △GHV. Prove: GD bisects ∠HGV. Ladder: 1. GH ≅ GV — Given; 2. GD is a median of △GHV — Given; 3. D is the midpoint of HV — Definition of median; 4. HD ≅ VD — Definition of midpoint; 5. GD ≅ GD — Reflexive property; 6. ? — SSS; 7. ∠HGD ≅ ∠VGD — CPCTC; 8. GD bisects ∠HGV — Definition of angle bisector. H V G D Not to scale - △GHD ≅ △GVD
- △GHV ≅ △GVD
- GD bisects ∠HGV
- △GHD ≅ △DVG
Tala shows that the median FY of pyramid face FCU splits its top angle evenly. Which statement fills the blank?
Sketch, not to scale: triangle FCU with F at the top and base CU; segment FY runs from F to point Y on CU. Marked: FC and FU with one tick each. Given: FC ≅ FU, and FY is a median of △FCU. Prove: FY bisects ∠CFU. Ladder: 1. FC ≅ FU — Given; 2. FY is a median of △FCU — Given; 3. Y is the midpoint of CU — Definition of median; 4. CY ≅ UY — Definition of midpoint; 5. FY ≅ FY — Reflexive property; 6. ? — SSS; 7. ∠CFY ≅ ∠UFY — CPCTC; 8. FY bisects ∠CFU — Definition of angle bisector. Sketch, not to scale: triangle FCU with F at the top and base CU; segment FY runs from F to point Y on CU. Marked: FC and FU with one tick each. Given: FC ≅ FU, and FY is a median of △FCU. Prove: FY bisects ∠CFU. Ladder: 1. FC ≅ FU — Given; 2. FY is a median of △FCU — Given; 3. Y is the midpoint of CU — Definition of median; 4. CY ≅ UY — Definition of midpoint; 5. FY ≅ FY — Reflexive property; 6. ? — SSS; 7. ∠CFY ≅ ∠UFY — CPCTC; 8. FY bisects ∠CFU — Definition of angle bisector. C U F Y Not to scale - FY bisects ∠CFU
- △FCY ≅ △YUF
- △FCU ≅ △FUY
- △FCY ≅ △FUY
Use this
Given: VD ≅ VG, and VC is a median of triangle VDG. Prove: VC bisects ∠DVG. Ladder: 1. VD ≅ VG -- Given; 2. VC is a median of triangle VDG -- Given; 3. C is the midpoint of DG -- Definition of median; 4. DC ≅ GC -- Definition of midpoint; 5. VC ≅ VC -- Reflexive property; 6. triangle VDC ≅ triangle VGC -- SSS; 7. ? -- ?; 8. VC bisects ∠DVG -- Definition of angle bisector.
Ari shows that the median VC of an isosceles ramp side VDG splits its top angle evenly. Which statement and reason fill the blanks?
Sketch, not to scale: triangle VDG with V at the top and base DG; segment VC runs from V to point C on DG. Marked: VD and VG with one tick each. Given: VD ≅ VG, and VC is a median of △VDG. Prove: VC bisects ∠DVG. Ladder: 1. VD ≅ VG — Given; 2. VC is a median of △VDG — Given; 3. C is the midpoint of DG — Definition of median; 4. DC ≅ GC — Definition of midpoint; 5. VC ≅ VC — Reflexive property; 6. △VDC ≅ △VGC — SSS; 7. ? — ?; 8. VC bisects ∠DVG — Definition of angle bisector. Sketch, not to scale: triangle VDG with V at the top and base DG; segment VC runs from V to point C on DG. Marked: VD and VG with one tick each. Given: VD ≅ VG, and VC is a median of △VDG. Prove: VC bisects ∠DVG. Ladder: 1. VD ≅ VG — Given; 2. VC is a median of △VDG — Given; 3. C is the midpoint of DG — Definition of median; 4. DC ≅ GC — Definition of midpoint; 5. VC ≅ VC — Reflexive property; 6. △VDC ≅ △VGC — SSS; 7. ? — ?; 8. VC bisects ∠DVG — Definition of angle bisector. V D C G Not to scale. - VC bisects ∠DVG — Definition of angle bisector
- ∠DVC ≅ ∠GVC — SSS
- ∠DVC ≅ ∠GVC — Reflexive property
- ∠DVC ≅ ∠GVC — CPCTC
Bo shows that the median UG of an isosceles pizza slice UCF splits its tip angle evenly. Which statement and reason fill the blanks?
Sketch, not to scale: triangle UCF with U at the top and base CF; segment UG runs from U to point G on CF. Marked: UC and UF with one tick each. Given: UC ≅ UF, and UG is a median of △UCF. Prove: UG bisects ∠CUF. Ladder: 1. UC ≅ UF — Given; 2. UG is a median of △UCF — Given; 3. G is the midpoint of CF — Definition of median; 4. CG ≅ FG — Definition of midpoint; 5. UG ≅ UG — Reflexive property; 6. ? — ?; 7. ∠CUG ≅ ∠FUG — CPCTC; 8. UG bisects ∠CUF — Definition of angle bisector. Sketch, not to scale: triangle UCF with U at the top and base CF; segment UG runs from U to point G on CF. Marked: UC and UF with one tick each. Given: UC ≅ UF, and UG is a median of △UCF. Prove: UG bisects ∠CUF. Ladder: 1. UC ≅ UF — Given; 2. UG is a median of △UCF — Given; 3. G is the midpoint of CF — Definition of median; 4. CG ≅ FG — Definition of midpoint; 5. UG ≅ UG — Reflexive property; 6. ? — ?; 7. ∠CUG ≅ ∠FUG — CPCTC; 8. UG bisects ∠CUF — Definition of angle bisector. C F U G Not to scale - △UCG ≅ △UFG — SSS
- △UCG ≅ △UFG — SAS
- △UCG ≅ △GFU — SSS
- ∠CUG ≅ ∠FUG — CPCTC
Esme shows that the median YW of an isosceles road sign YDS splits its top angle evenly. Which statement and reason fill the blanks?
Sketch, not to scale: triangle YDS with Y at the top and base DS; segment YW runs from Y to point W on DS. Marked: YD and YS with one tick each. Given: YD ≅ YS, and YW is a median of △YDS. Prove: YW bisects ∠DYS. Ladder: 1. YD ≅ YS — Given; 2. YW is a median of △YDS — Given; 3. W is the midpoint of DS — Definition of median; 4. DW ≅ SW — Definition of midpoint; 5. YW ≅ YW — Reflexive property; 6. ? — ?; 7. ∠DYW ≅ ∠SYW — CPCTC; 8. YW bisects ∠DYS — Definition of angle bisector. Sketch, not to scale: triangle YDS with Y at the top and base DS; segment YW runs from Y to point W on DS. Marked: YD and YS with one tick each. Given: YD ≅ YS, and YW is a median of △YDS. Prove: YW bisects ∠DYS. Ladder: 1. YD ≅ YS — Given; 2. YW is a median of △YDS — Given; 3. W is the midpoint of DS — Definition of median; 4. DW ≅ SW — Definition of midpoint; 5. YW ≅ YW — Reflexive property; 6. ? — ?; 7. ∠DYW ≅ ∠SYW — CPCTC; 8. YW bisects ∠DYS — Definition of angle bisector. D S Y W Not to scale - △YDW ≅ △YSW — SAS
- △YDW ≅ △YSW — SSS
- △YDW ≅ △WSY — SSS
- ∠DYW ≅ ∠SYW — CPCTC

