Question

prove that the line segment joining the midpoints of the hypotenuse of a right angled triangle to its opposite vertex is half of the hypotenuse.

Answer 1

Let △ABC△ABC be a right angled triangle, right angled at BB. Therefore the hypotenuse is ACAC. Let the middle point of ACAC be DD. Now draw a perpendicular DEDE on BABA and DFDF on CBCB. Now note that DE∥CBDE‖CB(converse of Theorem 3) and similarly DF∥AB⟹DF∥EBDF‖AB⟹DF‖EB(corollary from Theorem 2). Now in △ADE△ADE and △DFC△DFC we have1.DA=CD1.DA=CD, 2.∠AED=∠DFC2.∠AED=∠DFC and 3.∠EDA=∠FED3.∠EDA=∠FED (from Theorem 3). There the two triangles are congruent. Hence we get AE=DFAE=DF. But since in quadrilateral DEBFDEBF we get DF∥EBDF‖EB and DE∥FADE‖FA hence quadrilateral DEBFDEBF is a parallelogram and so DF=EBDF=EB and DE=FADE=FA thus giving us AE=EBAE=EBmaking EE the midpoint of BABA. Now we note that in △ABD△ABD we get thatDEDE is the perpendicular bisector of BABA. Hence △ABD△ABD is isosceles (converse of Theorem 8). We then get AD=DBAD=DB. Similarly by proving that △DBC△DBC is isosceles we get DB=CDDB=CD. Thus the theorem is proved.

Answer 2

Answer:

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