Answer:
Yes
Step-by-step explanation:
You can conclude that ΔGHI is congruent to ΔKJI, because you can see/interpret that there all the angles are congruent with one another, like with vertical angles (∠GIH and ∠KIJ) and alternate interior angles (∠H and ∠J, ∠G and ∠K).
We also know that we have two congruent sides, since it provides the information that line GK bisects line HJ, meaning that they have been split evenly (they have been split, with even/same lengths).
<u><em>So now we have three congruent angles, and two congruent sides. This is enough to prove that ΔGHI is congruent to ΔKJI,</em></u>
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The minimum number of sides that a polygon must have is 3
D) construction of the angle bisector.
So if you have a triangle that is not equilateral the perpendicular bisectors would not work so you need angle bisectors.
I have drawn a diagram
The green is the angle bisectors and the yellow the perpendicular bisectors.
You can see the problem straight away
Answer:
11. This is counting only odd numbers, or counting by two starting with one.
Step-by-step explanation:
