see if they are the same. If corresponding angles are then . Using a compass and straightedge copy 2DE onto and see if 2DE = If they are, then = by inspection. AC. DE 4. Prove that the medians of a triangle meet at a point. Suggested Proofs: Regular Geometry 1a / Honors Geometry 1a a) Proof by Midpoint (Formal Classic Approach) – This approach uses the proportional parts relationship (G.SRT.4) and properties of a parallelogram (G.CO.11). Therefore, this approach must wait to be presented until these concepts have been covered. Given: where and are medians. ∆ABC, Prove: That the 3rd median goes through point G. It is not unique that two intersect at a point, but it is very special when three do. That point is known as a point of concurrency. Create and mark where the ray meets as point Draw an E. auxiliary line parallel to median through and let it intersect C at H. is the midpoint of and ||, therefore is the midpoint of by F G proportional parts. is the midpoint of and is the midpoint of , thus is the D G midsegment of making || (or its extension). Therefore, || and || making quadrilateral a BHCG parallelogram. In any parallelogram, diagonals bisect, thus BE = making the midpoint of and a median concurrent with EC E the two medians. b) by Midsegment (Formal Classic Approach) – This approach uses properties of a parallelogram (G.CO.11). Therefore, this approach must wait to be presented until these concepts have been covered. where and are medians. ∆ABC, Prove: That the medians are concurrent. In is parallel and half the size of due to the ∆ABC, midsegment theorem. Create the midpoint of and , and J K respectively. In is parallel and half the size of due to ∆AGC, the midsegment theorem. Therefore, ||, by transitive, and , by substitution. Therefore, quadrilateral a DEKJ parallelogram, because one pair of opposite sides are both and . Since diagonals of a parallelogram bisect each other, = DG and = We already know that = and = GK JG GE. AJ JG CK
see if they are the same. If corresponding angles are then . Using a compass and straightedge copy 2DE onto and see if 2DE = AC. If they are, then DE = by inspection. 4. Prove that the medians of a triangle meet at a point. Suggested Proofs: Regular Geometry 1a / Honors Geometry 1a a) Proof by Midpoint (Formal – Classic Approach) This approach uses the proportional parts relationship (G.SRT.4) and properties of a parallelogram (G.CO.11). Therefore, this approach must wait to be presented until these concepts have been covered. Given: ∆ABC, where and are medians. Prove: That the 3rd median goes through point G. It is not unique that two intersect at a point, but it is very special when three do. That point is known as a point of concurrency. Create and mark where the ray meets as point E. Draw an auxiliary line parallel to median through C and let it intersect at H. F is the midpoint of and ||, therefore G is the midpoint of by proportional parts. D is the midpoint of and G is the midpoint of , thus is the midsegment of making || (or its extension). Therefore, || and || making quadrilateral BHCG a parallelogram. In any parallelogram, diagonals bisect, thus BE = EC making E the midpoint of and a median concurrent with the two medians. b) by Midsegment (Formal – Classic Approach) This approach uses properties of a parallelogram (G.CO.11). Therefore, this approach must wait to be presented until these concepts have been covered. ∆ABC, where and are medians. Prove: That the medians are concurrent. In ∆ABC, is parallel and half the size of due to the midsegment theorem. Create the midpoint of and , J and K respectively. In ∆AGC, is parallel and half the size of due to the midsegment theorem. Therefore, ||, by transitive, and , by substitution. Therefore, quadrilateral DEKJ a parallelogram, because one pair of opposite sides are both and . Since diagonals of a parallelogram bisect each other, DG = GK and JG = GE. We already know that AJ = JG and CK =
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