10 points! i need help filling this in for school asap!
1 answer:
Answer:
Given : EFGH is a parallelogram.
Prove: EG bisects HF and HF bisects EG.
Since, a parallelogram has two pairs of congruent and parallel sides.
Therefore, In parallelogram EFGH,
EF≅GH and EH≅FG
Also, EF║GH and EH║FG
Since,In triangles EKF and GKH,
EF≅HG
Since, EF║HG
⇒∠FEK ≅ ∠HGK and ∠EFK ≅ ∠ GHK ( When two parallel lines are cut by a transversal alternative interior angles are congruent )
⇒ Δ EKF ≅ Δ GKH ( ASA congruence postulate )
⇒ EK ≅ GK and FK ≅ HK ( CPCTC)
⇒ EG bisects HF and HF bisects EG ( Definition of bisector)
Hence proved.
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<em>Options:</em>
<em>Rotation of 180° about point B </em>
<em>Rotation of 90° about point b </em>
<em>Reflection over the x-axis </em>
<em>Translation down 2 units</em>
<em></em>
Answer:
Rotation of 180° about point B
<em></em>
Step-by-step explanation:
Considering coordinates of point B
Assume the coordinates of the line at point B is (x,y)
i.e.
First, we need to determine the slope at point B
Taking coordinates about the origin.
The slope of B is:
Where
--- origin
becomes
Taking the options one after the other:
Option A.
When rotated by 180°, the resulting coordinates of B would be
Taking the slope of B'
The slope of B is:
Where
--- origin
becomes
Notice that the slope of B and B' is the same;
<em>Hence:</em>
<em>Rotation of 180° about point B answers the question</em>
<em></em>
<em>There's no need to check for other options</em>