PLEASE! I URGENTLY NEED YOUR HELP! This problem is from my Advanced Geometry cla

Question

PLEASE! I URGENTLY NEED YOUR HELP! This problem is from my Advanced Geometry class:

a. Prove that AG = BH. That is, prove that the Saccheri quadrilateral associated to ABC is, in fact, Saccheri.

b. Prove that the line joining the midpoints of the two sides (not the base) of a triangle has length less than one-half the length of the base.

c. Prove that the sum of the interior angles of ABC is equal to the sum of the summit angles in its associated Saccheri quadrilateral (GAB and HBA in our picture).

Definition. Consider AABC. Let E and F be the midpoints of BC and AC, respectively, and let G and H be the points where the perpendiculars to EF from A and B intersect EF. We get The quadrilateral DABHG is said to be the Saccheri quadrilateral associated to AABC. The segment and point in red is my meta-hint for you. It is a good one.

Explanation / Answer

a. Suppose that the perpendicular from the vertex C of the CEF meets EF at J. Then, CJ and BH , being perpedicular to the same are GH , are parallel to each other. Similarly, CJ is parallel to AG also.Further, in the triangles CJF and BFH, the CJF = BHF = 900. Further, the s JCF and FBH being alternate s are equal. Further, CF = FB. Hence, the triangles CJF and BFH are congruent so that CJ = BH. Similarly, the triangles CJE and AGE are congruent so that CJ = AG. Hence AG = BH.

b. On comparing the triangles CEF , and C AB,we have ½ = CE/CA= CF/BC. Further, the C is common to both the s. Therefore, the CEF is silmilar to the C AB . Hence EF/AB = 1/2,. Thus, the line joining the midpoints of the two sides (not the base) of a triangle has length equal to ( not less than) one-half the length of the base.

c. We know that the sum of the four internal angles of a convex quadrilateral is 3600. Since each of the s AGH and BHG is a right angle, hence the sum of GAB and HBA = 3600 -1800 = 1800 = the sum of the internal angles of the ABC.

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