Congruence #4 - Q&A

Question 1 △ 𝐴 𝐵 𝐷 ≅ △ 𝐶 𝐵 𝐷 △ABD≅△CBD. Name the theorem or postulate that justifies the congruence. Options: A) AAS B) SSS C) ASA D) SAS Answer 1 Step 1 Consider the given figure. We need to identify the theorem or postulate that justifies △ 𝐴 𝐵 𝐷 ≅ △ 𝐶 𝐵 𝐷 △ABD≅△CBD. Given Information ∠ 𝐷 𝐴 𝐵 = ∠ 𝐷 𝐶 𝐵 ∠DAB=∠DCB ∠ 𝐴 𝐵 𝐷 = ∠ 𝐶 𝐵 𝐷 = 9 0 ∘ ∠ABD=∠CBD=90 ∘ BD is a common side. Step 2 By the AAS (Angle-Angle-Side) congruence postulate, we have enough information to conclude that: △ 𝐴 𝐵 𝐷 ≅ △ 𝐶 𝐵 𝐷 △ABD≅△CBD Answer: Option (A) AAS is correct. Question 2 Complete all statements and reasons for the following proof problem. Given: ∠ 𝑅 ∠R and ∠ 𝑉 ∠V are right angles. ∠ 1 ≅ ∠ 2 ∠1≅∠2 Prove: △ 𝑅 𝑆 𝑇 ≅ △ 𝑉 𝑆 𝑇 △RST≅△VST Answer 1 Step 1 Given: ∠ 𝑅 = ∠ 𝑉 = 9 0 ∘ ∠R=∠V=90 ∘ , ∠ 1 ≅ ∠ 2 ∠1≅∠2 For proving the given statement, we take triangles △ 𝑅 𝑆 𝑇 △RST and △ 𝑉 𝑆 𝑇 △VST. Step 2 So, in △ 𝑅 𝑆 𝑇 △RST and △ 𝑉 𝑆 𝑇 △VST: ∠ 𝑅 = ∠ 𝑉 ∠R=∠V (given) ∠ 1 = ∠ 2 ∠1=∠2 (given) 𝑆 𝑇 = 𝑆 𝑇 ST=ST (common side) Conclusion By the ASA (Angle-Side-Angle) congruence rule, we conclude: △ 𝑅 𝑆 𝑇 ≅ △ 𝑉 𝑆 𝑇 △RST≅△VST Hence, △ 𝑅 𝑆 𝑇 △RST and △ 𝑉 𝑆 𝑇 △VST are congruent. Question 3 Write a paragraph proof. Given: 𝐴 𝐷 AD and 𝐸 𝐶 EC intersect at 𝐵 B. Prove: △ 𝐴 𝐵 𝐸 ∼ △ 𝐷 𝐵 𝐶 △ABE∼△DBC Answer 1 Step 1: To Prove Given: 𝐴 𝐷 AD and 𝐸 𝐶 EC intersect at 𝐵 B. To determine: △ 𝐴 𝐵 𝐸 ∼ △ 𝐷 𝐵 𝐶 △ABE∼△DBC Step 2: Proof In △ 𝐴 𝐵 𝐸 △ABE and △ 𝐷 𝐵 𝐶 △DBC: 𝐴 𝐵 𝐷 𝐵 = 12 4 = 3 DB AB ​ = 4 12 ​ =3 𝐵 𝐸 𝐵 𝐶 = 15 5 = 3 BC BE ​ = 5 15 ​ =3 𝐴 𝐸 𝐷 𝐶 = 10 10 3 = 3 DC AE ​ = 3 10 ​ 10 ​ =3 Since the ratios of corresponding sides are equal, by the SSS similarity criterion, we have: △ 𝐴 𝐵 𝐸 ∼ △ 𝐷 𝐵 𝐶 △ABE∼△DBC Hence, it is proved. Question 4 To prove: If the convex quadrilateral 𝐴 𝐵 𝐶 𝐷 ABCD has opposite sides congruent, then the opposite sides are parallel. Given: The given figure is a quadrilateral. Answer 1 Approach: For proving the statement, use construction and congruency properties. Calculation: 𝐴 𝐵 ≅ 𝐷 𝐶 AB≅DC (Given) 𝐵 𝐶 ≅ 𝐴 𝐷 BC≅AD (Given) 𝐵 𝐷 ≅ 𝐵 𝐷 BD≅BD (Same side) Conclusion: Using the SSS (Side-Side-Side) congruence criterion, it can be said that: △ 𝐴 𝐵 𝐷 ≅ △ 𝐶 𝐷 𝐵 △ABD≅△CDB So, if △ 𝐴 𝐵 𝐷 ≅ △ 𝐶 𝐷 𝐵 △ABD≅△CDB, then ∠ 𝐶 𝐵 𝐷 ≅ ∠ 𝐴 𝐷 𝐵 ∠CBD≅∠ADB and ∠ 𝐴 𝐵 𝐷 ≅ ∠ 𝐶 𝐷 𝐵 ∠ABD≅∠CDB, which implies that alternate interior angles are equal. If alternate interior angles are equal, then the sides containing them are parallel. Therefore, 𝐵 𝐶 ∥ 𝐴 𝐷 BC∥AD and 𝐴 𝐵 ∥ 𝐶 𝐷 AB∥CD. Question 5 Find all whole number solutions of the congruence equation. ( 2 𝑥 + 1 ) ≡ 5 m o d     4 (2x+1)≡5mod4 Answer 1 Step 1 Take a look at the provided congruence equation. ( 2 𝑥 + 1 ) ≡ 5 m o d     4 (2x+1)≡5mod4 Step 2 Substitute each whole number less than four into the congruence equation: For 𝑥 = 0 x=0: 2 ( 0 ) + 1 = 1 ≠ 5 m o d     4 2(0)+1=1  =5mod4 (not a solution) For 𝑥 = 1 x=1: 2 ( 1 ) + 1 = 3 ≠ 5 m o d     4 2(1)+1=3  =5mod4 (not a solution) For 𝑥 = 2 x=2: 2 ( 2 ) + 1 = 5 ≡ 5 m o d     4 2(2)+1=5≡5mod4 (a solution) For 𝑥 = 3 x=3: 2 ( 3 ) + 1 = 7 ≠ 5 m o d     4 2(3)+1=7  =5mod4 (not a solution) Conclusion 0 and 2 are solutions within the modulo 4 cycle. Repeating this modulo pattern gives us the solutions as 𝑥 = 0 , 2 , 4 , 6 , 8 , 10 , … x=0,2,4,6,8,10,…. Final Answer: The congruence equation's solutions are therefore 0 , 2 , 4 , 6 , 8 , 10 , … 0,2,4,6,8,10,….