1) $\overline{AB} \cong \overline{BD}$ , $\overline{CD} \perp \overline{BD}$ , O is the midpoint of $\overline{BD}$ , $\overline{AB} \cong \overline{CD}$ (given)

2) $\angle ABO, \angle ODC$ are right angles (perpendicular lines form right angles)

3) $\triangle ABO, \triangle CDO$ are right triangles (a triangle with a right angle is a right triangle)

4) $\overline{BO} \cong \overline{OD}$ (a midpoint splits a segment into two congruent parts)

5) $\triangle ABO \cong \triangle CDO$ (LL)

Question 7

1) $\angle ADC, \angle BDC$ are right angles), $\overline{AD} \cong \overline{BD}$

2) $\overline{CD} \cong \overline{CD}$ (reflexive property)

3) $\triangle CDA, \triangle CDB$ are right triangles (a triangle with a right angle is a right triangle)

4) $\triangle ADC \cong \triangle BDC$ (LL)

5) $\overline{AC} \cong \overline{BC}$ (CPCTC)

Question 8

1) $\overline{CD} \perp \overline{AB}$ , point D bisects $\overline{AB}$ (given)

2) $\angle CDA, \angle CDB$ are right angles (perpendicular lines form right angles)

3) $\triangle CDA, \triangle CDB$ are right triangles (a triangle with a right angle is a right triangle)

4) $\overline{AD} \cong \overline{DB}$ (definition of a bisector)

5) $\overline{CD} \cong \overline{CD}$ (reflexive property)

6) $\triangle ADC \cong \triangle BDC$ (LL)

7) $\angle ACD \cong \angle BCD$ (CPCTC)

8 0

1 year ago

Emily thought of a counting number, and then she added the next counting number to it. She got the sum of 57. What was Emily’s n

Dovator [93]

Let this number be x. The next number is (x+1). And their sum is 57. Then, we have to solve this equation x+x+1=57
$2x=56$
$x=28$
Emily's number was 28

8 0

3 years ago

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