Given AD = BC and AD parallel BC Prove ABCD is a parallelogram proof

Mathematics

2 answers:

Angelina_Jolie [31]3 years ago

7 0

Statements Reasons

1. AD ≅ BC; AD ∥ BC 1. given

2. ∠CAD and ∠ACB are alternate interior ∠s 2. definition of alternate interior angles

3. ∠CAD ≅ ∠ACB 3. alternate interior angles are congruent

4. AC ≅ AC 4. reflexive property

5. △CAD ≅ △ACB 5. SAS congruency theorem

6. AB ≅ CD 6. Corresponding Parts of Congruent triangles are Congruent (CPCTC)

7. ABCD is a parallelogram 7. parallelogram side theorem

natka813 [3]3 years ago

5 0

Answer: the ABC is up illogic of the Paul Graham that makes the box square root to DC=BA

Step-by-step explanation:

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3.a. Solve the following equations.<br>

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Answer:

1. Solving $\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)$ we get x=3

2. Solving $\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}$ we get x=-1

Step-by-step explanation:

We need to solve the equations:

1. $\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)$

Expanding the brackets

$\frac{2x}{3}-\frac{1}{3} -\frac{2x}{4}-\frac{1}{4} =\frac{2}{12}-\frac{x}{12} \\\frac{2x}{3}-\frac{1}{3} -\frac{x}{2}-\frac{1}{4} =\frac{1}{6}-\frac{x}{12}$

Taking LCM on left side we get 12

$\frac{2x*4-1*4-x*6-1*3}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{8x-4-6x-3}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{8x-6x-3-4}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{2x-7}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{2x}{12}-\frac{7}{12}= \frac{1}{6}-\frac{x}{12}$

Add 7/12 on both sides

$\frac{2x}{12}-\frac{7}{12}+\frac{7}{12} = \frac{1}{6}-\frac{x}{12}+\frac{7}{12}\\\frac{x}{6}=\frac{1*2-x+7}{12}\\ \frac{x}{6}=\frac{-x+9}{12} \\\frac{x}{6}=-\frac{x}{12} +\frac{9}{12}\\ \frac{x}{6}=-\frac{x}{12} +\frac{3}{4}$