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

Circle the best answer.
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Given: ∠DCF is a right angle, and CQ ⊥ DF. Prove: FC² = FQ * FD. Ladder: 1. ∠DCF is a right angle, and CQ ⊥ DF -- Given; 2. ∠FQC is a right angle -- Definition of perpendicular lines; 3. ∠FQC ≅ ∠FCD -- Right angles congruence theorem; 4. ∠QFC ≅ ∠CFD -- Reflexive property; 5. triangle FQC ~ triangle FCD -- AA similarity postulate; 6. FQ/FC = FC/FD -- Definition of similar triangles; 7. FC² = FQ * FD -- ?.
Esme uses the same ramp CDF and support CQ, now working from corner F. Which reason fills the blank?
- Given
- Division property of equality
- Cross products property
- Pythagorean theorem
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Given: ∠USW is a right angle, and SY ⊥ UW. Prove: US² = UY * UW. Ladder: 1. ∠USW is a right angle, and SY ⊥ UW -- Given; 2. ∠UYS is a right angle -- Definition of perpendicular lines; 3. ∠UYS ≅ ∠USW -- Right angles congruence theorem; 4. ∠YUS ≅ ∠SUW -- Reflexive property; 5. triangle UYS ~ triangle USW -- ?; 6. UY/US = US/UW -- Definition of similar triangles; 7. US² = UY * UW -- Cross products property.
Jun draws a tent side as right triangle SUW, with a pole SY to the slope UW. Which reason fills the blank?
Sketch, not to scale: right triangle SUW with the right angle at S and hypotenuse UW; segment SY runs from S to point Y on UW. Marked: right-angle boxes at S and Y. Given: ∠USW is a right angle, and SY ⊥ UW. Prove: US² = UY · UW. Ladder: 1. ∠USW is a right angle, and SY ⊥ UW — Given; 2. ∠UYS is a right angle — Definition of perpendicular lines; 3. ∠UYS ≅ ∠USW — Right angles congruence theorem; 4. ∠YUS ≅ ∠SUW — Reflexive property; 5. △UYS ~ △USW — ?; 6. UY/US = US/UW — Definition of similar triangles; 7. US² = UY · UW — Cross products property. Sketch, not to scale: right triangle SUW with the right angle at S and hypotenuse UW; segment SY runs from S to point Y on UW. Marked: right-angle boxes at S and Y. Given: ∠USW is a right angle, and SY ⊥ UW. Prove: US² = UY · UW. Ladder: 1. ∠USW is a right angle, and SY ⊥ UW — Given; 2. ∠UYS is a right angle — Definition of perpendicular lines; 3. ∠UYS ≅ ∠USW — Right angles congruence theorem; 4. ∠YUS ≅ ∠SUW — Reflexive property; 5. △UYS ~ △USW — ?; 6. UY/US = US/UW — Definition of similar triangles; 7. US² = UY · UW — Cross products property. Sketch, not to scale S U W Y - AA similarity postulate
- ASA
- HL
- SAS similarity theorem
Use this
Given: ∠USW is a right angle, and SY ⊥ UW. Prove: WS² = WY * WU. Ladder: 1. ∠USW is a right angle, and SY ⊥ UW -- Given; 2. ∠WYS is a right angle -- Definition of perpendicular lines; 3. ∠WYS ≅ ∠WSU -- Right angles congruence theorem; 4. ∠YWS ≅ ∠SWU -- Reflexive property; 5. triangle WYS ~ triangle WSU -- ?; 6. WY/WS = WS/WU -- Definition of similar triangles; 7. WS² = WY * WU -- Cross products property.
Kira uses the same tent side SUW and pole SY, now working from corner W. Which reason fills the blank?
Sketch, not to scale: right triangle SUW with the right angle at S and hypotenuse UW; segment SY runs from S to point Y on UW. Marked: right-angle boxes at S and Y. Given: ∠USW is a right angle, and SY ⊥ UW. Prove: WS² = WY · WU. Ladder: 1. ∠USW is a right angle, and SY ⊥ UW — Given; 2. ∠WYS is a right angle — Definition of perpendicular lines; 3. ∠WYS ≅ ∠WSU — Right angles congruence theorem; 4. ∠YWS ≅ ∠SWU — Reflexive property; 5. △WYS ~ △WSU — ?; 6. WY/WS = WS/WU — Definition of similar triangles; 7. WS² = WY · WU — Cross products property. Sketch, not to scale: right triangle SUW with the right angle at S and hypotenuse UW; segment SY runs from S to point Y on UW. Marked: right-angle boxes at S and Y. Given: ∠USW is a right angle, and SY ⊥ UW. Prove: WS² = WY · WU. Ladder: 1. ∠USW is a right angle, and SY ⊥ UW — Given; 2. ∠WYS is a right angle — Definition of perpendicular lines; 3. ∠WYS ≅ ∠WSU — Right angles congruence theorem; 4. ∠YWS ≅ ∠SWU — Reflexive property; 5. △WYS ~ △WSU — ?; 6. WY/WS = WS/WU — Definition of similar triangles; 7. WS² = WY · WU — Cross products property. hypotenuse UW S W U Y Not to scale. - HL
- SSS similarity theorem
- AAS
- AA similarity postulate
Luz proves the Pythagorean theorem with similar triangles on a ramp HQU. Which statement fills the blank?
Sketch, not to scale: right triangle HQU with the right angle at H and hypotenuse QU; segment HF runs from H to point F on QU. Marked: right-angle boxes at H and F. Given: QH² = QF · QU and UH² = UF · QU, from the similar triangles. Prove: QH² + UH² = QU². Ladder: 1. QH² = QF · QU — Given; 2. UH² = UF · QU — Given; 3. QH² + UH² = QF · QU + UF · QU — Addition property of equality; 4. ? — Distributive property; 5. QF + UF = QU — Segment addition postulate; 6. QH² + UH² = QU² — Substitution property. Sketch, not to scale: right triangle HQU with the right angle at H and hypotenuse QU; segment HF runs from H to point F on QU. Marked: right-angle boxes at H and F. Given: QH² = QF · QU and UH² = UF · QU, from the similar triangles. Prove: QH² + UH² = QU². Ladder: 1. QH² = QF · QU — Given; 2. UH² = UF · QU — Given; 3. QH² + UH² = QF · QU + UF · QU — Addition property of equality; 4. ? — Distributive property; 5. QF + UF = QU — Segment addition postulate; 6. QH² + UH² = QU² — Substitution property. H Q U F Not to scale - QH² + UH² = QU²
- QH² + UH² = QU(QF + UF)
- QH + UH = QU
- QH² + UH² = QU + QF + UF
Milo proves the Pythagorean theorem with similar triangles on a roof corner WGY. Which statement fills the blank?
Sketch, not to scale: right triangle WGY with the right angle at W and hypotenuse GY; segment WC runs from W to point C on GY. Marked: right-angle boxes at W and C. Given: GW² = GC · GY and YW² = YC · GY, from the similar triangles. Prove: GW² + YW² = GY². Ladder: 1. GW² = GC · GY — Given; 2. YW² = YC · GY — Given; 3. GW² + YW² = GC · GY + YC · GY — Addition property of equality; 4. GW² + YW² = GY(GC + YC) — Distributive property; 5. ? — Segment addition postulate; 6. GW² + YW² = GY² — Substitution property. Sketch, not to scale: right triangle WGY with the right angle at W and hypotenuse GY; segment WC runs from W to point C on GY. Marked: right-angle boxes at W and C. Given: GW² = GC · GY and YW² = YC · GY, from the similar triangles. Prove: GW² + YW² = GY². Ladder: 1. GW² = GC · GY — Given; 2. YW² = YC · GY — Given; 3. GW² + YW² = GC · GY + YC · GY — Addition property of equality; 4. GW² + YW² = GY(GC + YC) — Distributive property; 5. ? — Segment addition postulate; 6. GW² + YW² = GY² — Substitution property. W G Y C Not to scale - GW² + YW² = GY²
- GC + YC = GY
- GC + GW = GY
- GC = YC
Noah proves the Pythagorean theorem with similar triangles on a road sign VSD. Which statement fills the blank?
Sketch, not to scale: right triangle VSD with the right angle at V and hypotenuse SD; segment VH runs from V to point H on SD. Marked: right-angle boxes at V and H. Given: SV² = SH · SD and DV² = DH · SD, from the similar triangles. Prove: SV² + DV² = SD². Ladder: 1. SV² = SH · SD — Given; 2. DV² = DH · SD — Given; 3. ? — Addition property of equality; 4. SV² + DV² = SD(SH + DH) — Distributive property; 5. SH + DH = SD — Segment addition postulate; 6. SV² + DV² = SD² — Substitution property. Sketch, not to scale: right triangle VSD with the right angle at V and hypotenuse SD; segment VH runs from V to point H on SD. Marked: right-angle boxes at V and H. Given: SV² = SH · SD and DV² = DH · SD, from the similar triangles. Prove: SV² + DV² = SD². Ladder: 1. SV² = SH · SD — Given; 2. DV² = DH · SD — Given; 3. ? — Addition property of equality; 4. SV² + DV² = SD(SH + DH) — Distributive property; 5. SH + DH = SD — Segment addition postulate; 6. SV² + DV² = SD² — Substitution property. V S D H Not to scale - SV² + DV² = SH · SD
- SV² = DV²
- SV² + DV² = SD²
- SV² + DV² = SH · SD + DH · SD
Pia proves the Pythagorean theorem with similar triangles on a pyramid face UFQ. Which statement fills the blank?
Sketch, not to scale: right triangle UFQ with the right angle at U and hypotenuse FQ; segment UG runs from U to point G on FQ. Marked: right-angle boxes at U and G. Given: FU² = FG · FQ and QU² = QG · FQ, from the similar triangles. Prove: FU² + QU² = FQ². Ladder: 1. FU² = FG · FQ — Given; 2. QU² = QG · FQ — Given; 3. FU² + QU² = FG · FQ + QG · FQ — Addition property of equality; 4. FU² + QU² = FQ(FG + QG) — Distributive property; 5. ? — Segment addition postulate; 6. FU² + QU² = FQ² — Substitution property. Sketch, not to scale: right triangle UFQ with the right angle at U and hypotenuse FQ; segment UG runs from U to point G on FQ. Marked: right-angle boxes at U and G. Given: FU² = FG · FQ and QU² = QG · FQ, from the similar triangles. Prove: FU² + QU² = FQ². Ladder: 1. FU² = FG · FQ — Given; 2. QU² = QG · FQ — Given; 3. FU² + QU² = FG · FQ + QG · FQ — Addition property of equality; 4. FU² + QU² = FQ(FG + QG) — Distributive property; 5. ? — Segment addition postulate; 6. FU² + QU² = FQ² — Substitution property. U F Q G Not to scale - FU² + QU² = FQ²
- FG = QG
- FG + FU = FQ
- FG + QG = FQ
Tala proves the Pythagorean theorem with similar triangles on a tent side FYH. Which statement and reason fill the blanks?
Sketch, not to scale: right triangle FYH with the right angle at F and hypotenuse YH; segment FV runs from F to point V on YH. Marked: right-angle boxes at F and V. Given: YF² = YV · YH and HF² = HV · YH, from the similar triangles. Prove: YF² + HF² = YH². Ladder: 1. YF² = YV · YH — Given; 2. HF² = HV · YH — Given; 3. YF² + HF² = YV · YH + HV · YH — Addition property of equality; 4. YF² + HF² = YH(YV + HV) — Distributive property; 5. YV + HV = YH — Segment addition postulate; 6. ? — ?. Sketch, not to scale: right triangle FYH with the right angle at F and hypotenuse YH; segment FV runs from F to point V on YH. Marked: right-angle boxes at F and V. Given: YF² = YV · YH and HF² = HV · YH, from the similar triangles. Prove: YF² + HF² = YH². Ladder: 1. YF² = YV · YH — Given; 2. HF² = HV · YH — Given; 3. YF² + HF² = YV · YH + HV · YH — Addition property of equality; 4. YF² + HF² = YH(YV + HV) — Distributive property; 5. YV + HV = YH — Segment addition postulate; 6. ? — ?. F Y H V Not to scale - YF² + HF² = YH² — Substitution property
- YF² + HF² = YH² — Pythagorean theorem
- YF² + HF² = YH² — Addition property of equality
- YF + HF = YH — Substitution property
Use this
Ladder: 1. ∠HGV is a right angle, and GD ⊥ HV -- Given; 2. ∠HDG is a right angle -- Definition of perpendicular lines; 3. ∠HDG ≅ ∠HGV -- Right angles congruence theorem; 4. ∠DHG ≅ ∠GHV -- Reflexive property; 5. triangle HDG ~ triangle HGV -- AA similarity postulate; 6. ? -- ?; 7. HG² = HD * HV -- Cross products property.
Zee draws a roof corner as right triangle GHV, with a beam GD to the slope HV. Which statement and reason fill the blanks?
Sketch, not to scale: right triangle GHV with the right angle at G and hypotenuse HV; segment GD runs from G to point D on HV. Marked: right-angle boxes at G and D. Given: ∠HGV is a right angle, and GD ⊥ HV. Prove: HG² = HD · HV. Ladder: 1. ∠HGV is a right angle, and GD ⊥ HV — Given; 2. ∠HDG is a right angle — Definition of perpendicular lines; 3. ∠HDG ≅ ∠HGV — Right angles congruence theorem; 4. ∠DHG ≅ ∠GHV — Reflexive property; 5. △HDG ~ △HGV — AA similarity postulate; 6. ? — ?; 7. HG² = HD · HV — Cross products property. Sketch, not to scale: right triangle GHV with the right angle at G and hypotenuse HV; segment GD runs from G to point D on HV. Marked: right-angle boxes at G and D. Given: ∠HGV is a right angle, and GD ⊥ HV. Prove: HG² = HD · HV. Ladder: 1. ∠HGV is a right angle, and GD ⊥ HV — Given; 2. ∠HDG is a right angle — Definition of perpendicular lines; 3. ∠HDG ≅ ∠HGV — Right angles congruence theorem; 4. ∠DHG ≅ ∠GHV — Reflexive property; 5. △HDG ~ △HGV — AA similarity postulate; 6. ? — ?; 7. HG² = HD · HV — Cross products property. G H V D Not to scale. - HD/HG = GD/HV — Definition of similar triangles
- HD/HG = HV/HG — Definition of similar triangles
- HD/HG = HG/HV — AA similarity postulate
- HD/HG = HG/HV — Definition of similar triangles
Use this
Given: ∠CYS is a right angle, and YW ⊥ CS. Prove: SY² = SW * SC. Ladder: 1. ∠CYS is a right angle, and YW ⊥ CS -- Given; 2. ∠SWY is a right angle -- Definition of perpendicular lines; 3. ∠SWY ≅ ∠SYC -- Right angles congruence theorem; 4. ∠WSY ≅ ∠YSC -- Reflexive property; 5. triangle SWY ~ triangle SYC -- AA similarity postulate; 6. ? -- ?; 7. SY² = SW * SC -- Cross products property.
Ari cuts a right-triangle sandwich YCS, then cuts from Y to the long edge CS at a right angle, meeting it at W. Which statement and reason fill the blanks?
Sketch, not to scale: right triangle YCS with the right angle at Y and hypotenuse CS; segment YW runs from Y to point W on CS. Marked: right-angle boxes at Y and W. Given: ∠CYS is a right angle, and YW ⊥ CS. Prove: SY² = SW · SC. Ladder: 1. ∠CYS is a right angle, and YW ⊥ CS — Given; 2. ∠SWY is a right angle — Definition of perpendicular lines; 3. ∠SWY ≅ ∠SYC — Right angles congruence theorem; 4. ∠WSY ≅ ∠YSC — Reflexive property; 5. △SWY ~ △SYC — AA similarity postulate; 6. ? — ?; 7. SY² = SW · SC — Cross products property. Sketch, not to scale: right triangle YCS with the right angle at Y and hypotenuse CS; segment YW runs from Y to point W on CS. Marked: right-angle boxes at Y and W. Given: ∠CYS is a right angle, and YW ⊥ CS. Prove: SY² = SW · SC. Ladder: 1. ∠CYS is a right angle, and YW ⊥ CS — Given; 2. ∠SWY is a right angle — Definition of perpendicular lines; 3. ∠SWY ≅ ∠SYC — Right angles congruence theorem; 4. ∠WSY ≅ ∠YSC — Reflexive property; 5. △SWY ~ △SYC — AA similarity postulate; 6. ? — ?; 7. SY² = SW · SC — Cross products property. Y C S W Not to scale. - SW/SY = SC/SY — Definition of similar triangles
- SW/SY = SY/SC — AA similarity postulate
- SW/SY = YW/SC — Definition of similar triangles
- SW/SY = SY/SC — Definition of similar triangles