10th grade Use congruent triangles to find missing sides and angles worksheet

Use congruent triangles to find missing sides and angles

Grade 10 Math Worksheet - Challenge Me

Circle the best answer.

1. 1.

   Ari proves E is the midpoint of BD. Which chain of steps is correct?

   • △ABC ≅ △ADC (SSS); ∠BAE = ∠DAE (CPCTC); △ABE ≅ △ADE (SAS); BE = DE (CPCTC)
   • △ABC ≅ △ADC (SAS); ∠BAE = ∠DAE (CPCTC); BE = DE
   • AC bisects ∠BAD in the picture; △ABE ≅ △ADE (SAS); BE = DE
   • AE = AE; AB = AD; △ABE ≅ △ADE (SSS), since E is on BD; BE = DE (CPCTC)
2. 2.

   Bo proves AE = CE. Which chain of steps is correct?

   • AB = DC; alternate angles give ASA, so △ABE ≅ △CDE; AE = CE by CPCTC
   • Assume E is the midpoint of both diagonals; then △ABE ≅ △CDE by SSS
   • AB = DC; ∠BAE = ∠DCE; SAS gives △ABE ≅ △CDE; AE = CE (CPCTC)
   • AB = DC; alternate angles; ASA gives △ABE ≅ △CED; AE = CE (CPCTC)
3. 3.

   Esme proves ∠B = ∠C. Which argument is correct?

   • AB = AC, ∠BAD = ∠CAD, AD = AD; SAS gives △ABD ≅ △ACD; ∠B = ∠C (CPCTC)
   • AB = AC and AD = AD; SSA gives △ABD ≅ △ACD; ∠B = ∠C by CPCTC
   • AD ⊥ BC in the picture; HL gives △ABD ≅ △ACD; ∠B = ∠C
   • Assume ∠B = ∠C; then AAS gives △ABD ≅ △ACD, so ∠B = ∠C
4. 4.

   Jun proves AC = DB. Which argument is correct?

   • BC = CB (shared); SAS gives △ABC ≅ △DCB; AC = DB by CPCTC
   • AC = DB is assumed, so SSS gives △ABC ≅ △DCB
   • Not enough is given: the triangles share no marked side or angle
   • BC = BC; SAS gives △ABC ≅ △DBC; AC = DC by CPCTC
5. 5.

   Quadrilateral ABCD with AB ∥ DC and AB = DC. Luz lists claims that must be true for the card. Which claim follows?

   • AD = BC and AD ∥ BC (△ABC ≅ △CDA by SAS, then CPCTC)
   • ∠A = 90°, since the quadrilateral looks like a rectangle
   • △ABC ≅ △ADC by SSS, so AB = AD
   • AC = BD, since parallelograms have equal diagonals
6. 6.

   Milo lists claims that must be true for the card. Which claim follows?

   • D is the midpoint of BC, and AD is perpendicular to BC
   • ∠B = ∠BAD, since AD bisects ∠BAC
   • D is the midpoint of BC, but AD ⊥ BC needs more information
   • AD = BD, as the picture suggests
7. 7.

   Noah lists claims that must be true for the card. Which claim follows?

   • AB = DC and AB ∥ DC (△ABM ≅ △DCM by SAS, then CPCTC)
   • AD = BC, since the diagonals of ABDC are equal in length
   • AB = DB, because the triangles are congruent
   • ∠BAC = 90°, since ABDC looks like a rectangle

8. Read this first

   Proof

   1. AB = DE (given). 2. BC = EF (given). 3. ∠A = ∠D (given). 4. △ABC ≅ △DEF (SAS). 5. AC = DF (CPCTC). The marked angles, ∠A and ∠D, are opposite BC and EF.

9. 8.

   Ren checks this proof. Which line is wrong, and why?

   • Line 4: the triangles should be named △ABC ≅ △FED
   • Line 4: ∠A is not between AB and BC, so SAS fails
   • Line 3: the angle cannot be given without a figure
   • Line 5: CPCTC cannot give sides equal

10. Use this

    Kite ABCD with AB = AD and CB = CD. Diagonals meet at E. Proof: 1. AB = AD, CB = CD (given). 2. AE = AE (reflexive). 3. BE = DE (shown). 4. △ABE ≅ △ADE (SSS). 5. ∠AEB = ∠AED (CPCTC). 6. Each is 90° (linear pair).

11. 9.

    Tala proves AC ⊥ BD. Which line is the gap?

    • Line 5: CPCTC cannot follow SSS
    • Line 3: BE = DE is not yet shown
    • Line 2: AE = AE needs a reason first
    • Line 4: it should be SAS, not SSS

12. Use this

    Proof: 1. AM = DM and BM = CM (midpoints). 2. ∠AMB = ∠DMC (vertical). 3. triangle AMB ≅ triangle DMC (SAS). 4. ∠BAM = ∠CDM (CPCTC). 5. AB ∥ DC.

13. 10.

    Zee checks the proof of AB ∥ DC. What is missing at line 5?

    • Line 1 is wrong: BM = CM is not given
    • Line 4 is wrong: CPCTC cannot give angles, just sides
    • Line 3 should be ASA, since an angle is used
    • A reason: equal alternate angles give parallel lines