Answer:
<u>Given rhombus ABCD with</u>
- m∠EAD = 67°, CE = 5, DE = 12
<u>Properties of a rhombus:</u>
- All sides are congruent
- Diagonals are perpendicular
- Diagonals are angle bisectors
- Diagonals bisect each other
<u>Solution, considering the above properties</u>
- 1. m∠AED = 90°, as angle between diagonals
- 2. m∠ADE = 90° - 67° = 23° as complementary of ∠EAD
- 3. m∠BAE = 67°, as ∠BAE ≅ ∠EAD
- 4. AE = CE = 5, as E is midpoint of AC
- 5. BE = DE = 12, as E is midpoint of BD
Answer
The LCM for 4xy², 2x²y3 is the numeric part 4 multiplied by the variable part.
4xy²
Step-by-step explanation:
Answer:
Part a) The radii are segments AC and AD and the tangents are the segments CE and DE
Part b) 
Step-by-step explanation:
Part a)
we know that
A <u>radius</u> is a line from any point on the circumference to the center of the circle
A <u>tangent</u> to a circle is a straight line which touches the circle at only one point. The tangent to a circle is perpendicular to the radius at the point of tangency.
In this problem
The radii are the segments AC and AD
The tangents are the segments CE and DE
Part b)
we know that
radius AC is perpendicular to the tangent CE
radius AD is perpendicular to the tangent DE
CE=DE
Triangle ACE is congruent with triangle ADE
Applying the Pythagoras Theorem

substitute the values and solve for CE





remember that
CE=DE
so
