In the triangle ABC, ∠ a > 90 °, BD and CE are the heights of the triangle respectively, and M is the midpoint of the edge BC, connecting de and DM, (1) complete the drawing according to the above requirements (2) To prove that △ MDE is an isosceles triangle (3) try to explore whether △ MDE can be a right triangle? If yes, please tell us the degree of ∠ BAC

In the triangle ABC, ∠ a > 90 °, BD and CE are the heights of the triangle respectively, and M is the midpoint of the edge BC, connecting de and DM, (1) complete the drawing according to the above requirements (2) To prove that △ MDE is an isosceles triangle (3) try to explore whether △ MDE can be a right triangle? If yes, please tell us the degree of ∠ BAC

2) Because point m is the midpoint of BC, triangle BCD and triangle BCE are right triangles, so DM = 0.5bc, EM = 0.5bc, so DM = De, so triangle DEM is isosceles triangle,
3) When angle a = 135 degrees, triangle DEM is right triangle, angle DME = 90 degrees

As shown in the figure, take sides AB and AC of triangle ABC as sides to make square ABDE and square acfg outward respectively, connect eg, BC and h as the midpoint of FG, ha intersects BC with m, proving that am is vertical to BC

Extend ah to Q, make HQ = ah, connect QE and QG,
Then the quadrilateral eagq is a parallelogram (if the diagonal is bisected, it is a parallelogram),
EQ = AG, (the opposite side is equal),
AG=AC,
EQ=AC,
EA=AB,
∵EQ//AG,
(QEA + EAG = 180 degrees,
BAC = 360-90-90 - < EAG = 180 - < EAG,
∴〈QEA=〈BAC,
∴△QEA≌△CAB,（SAS）
∴〈ABC=〈EAQ,
"EAQ +" EAB + "BAM = 180 degrees,
EAB = 90 degrees,
EAQ + BAM = 90 degrees,
MBA + BAM = 90 degrees,
BMA = 180 degrees - MBA - BAM = 180 degrees - 90 degrees = 90 degrees,
∴AM⊥BC,

As shown in Figure 15-1, make square abgf and square ACDE with AB and AC of △ ABC as sides, connect Ag and ad, which is the midpoint of Ag, M is the midpoint of AD, and K is the middle of BC
Taking the edges AB and AC of △ ABC as the edges to make square abgf and square ACDE respectively, and connecting Ag and ad, n is the midpoint of Ag, M is the midpoint of AD, K is the midpoint of BC, MK and NK relations (two!

It is proved that: to connect FB, EC, FC and EB, because AF = AB, AC = AE ∠ fac = ∠ BAE = 90 ° + ∠ BAC ≠ △ fac & # 8773; △ BAE (SAS) ≠ FC = be, because FN = Nb, BK = KC ≠ NK ‖ = FC / 2, similarly: MK ‖ = be / 2 ‖ NK = MK, let FC intersect be at P, MK at R, be at NK at h, AC at Q, prkh is parallelogram

20. As shown in the figure, take the sides AB and AC of △ ABC as the sides to make square abgf and square ACDE, connect Ag and ad. n is the midpoint of Ag, M is the midpoint of AD, and K is BC

Kmn is RT △ and K is right angle
Then be / / km, CF / / kn, we prove that be is perpendicular to CF
The congruent △ EAB of △ CDF is easily obtained by angular substitution
If HCE + HEC = 90 degrees, EHC = 90 degrees
(1) (2) easier to prove