Proof: As shown in the figure, DE is a Simson line, which connects and extends AH and intersects at point F; Make ray MG, make ∠FMG=∠KAM, and the intersection line AH is at point G; Let MS be parallel to BC, AH parallel to s, MP and BC intersect at point n, MK and AH intersect at point L, AF and BC intersect at point T, AQ and KM, and BC intersect at point X and Y respectively. Connect PB, PD, PE, AQ, KN, AK, AM, CM, CH.
∵PE⊥AB,PD⊥BC
∴PBED*** ring
∴∠aed =∠bpd = 90-∠PBC = 90-( 1/2)arc PC = 90-( 1/2)arc bq = 90-∠baq。
Namely DE⊥AQ
MK∨DE again
∴MK⊥AQ
∵PQKM***
∴∠ QKM+∠ QPM = ∠ JKM+∠ JNM =180, that is, NJKM*** cycle.
∴∠JKM=∠MNC,∠KMJ=∠KNJ
So it is necessary to prove that △KMJ is an isosceles triangle, or ∠JKM=∠KMJ.
Just prove that ∠MNC=∠KNJ.
Note that H is the vertical center, so H and F are symmetrical about BC (this is easy to prove, so I won't elaborate here).
Therefore, only the KNF*** line is needed.
Let's prove KNF*** line by Menelius theorem, and take △MLH.
Prove KNF*** line
Just prove that (MK/KL) (LF/FH) (HN/nm) =1(1).
And HN/nm = s (△ CNH)/s (△ CNM) = cn (1/2) HF/(cnst) = (1/2) HF/ST.
MK/KL = S(△AKM)/S(△AKL)= AM AK sin∠KAM/(AK ALsin∠KAL)= AM sin∠KAM/(ALsin∠KAL)
Substituting (1), we only need to prove that (am lf sin ∠ kam)/(al st sin ∠ kal) = 2 (2).
And lf/al = s (△ LFM)/s (△ LMA) = mfsin ∠ fmk/(amsin ∠ amk).
And ∠FMK =∠ Karl.
Substituting equation (2), we only need to prove that (mfsin ∠ kam)/(st sin ∠ amk) = 2?
Or MF km/(ST AK) = 2? (3)
On the other hand, FMG=∠KAM and GFM=∠MKA.
∴△GFM∽△MKA
∴KM/KA=FG/FM？ (4)
∠G=∠AML
Also pay attention to LXYT*** cycles (AQ⊥AM, AF⊥BC) and AQPM*** cycles.
∴∠AMH+∠AQP=∠AMH+∠AYN= 180，∠XLT+∠XYT=∠ALM+∠AYN= 180
∴∠AMH=∠ALM
∴∠AML=∠AHM=∠G, that is, △MGH is an isosceles triangle.
So GF=GH+HF=2(SH+HT)=2ST? (5)
Substituting (4)(5) into (3) can prove (3).
This proves KNF's * * * route.
So △KMJ is an isosceles triangle.