Question

已知,BE,CF是锐角三角形ABC的高,在BE上截取BP=AC在CF,在CF上截取CQ=AB,PM⊥BC于M,QN⊥BC

Answer 1

过A作AD⊥BC于D

BP=AC,

∠PMB=∠ADC

∠EBC+∠BCE=∠DAC+∠ACD=90°

∠EBC=∠DAC

BP=AC

△ADC≌△BDP

PM=DC

同理∠BAD+∠ABC=∠QCN+∠ABC=90°

∠BAD=∠QCN,直角,CQ=AB

△ADB≌△CNQ

QN=BD

BD+DC=BC

PM+QN=BC

若△ABC中∠A为钝角,同样方法得PM+QN=BC

若△ABC中∠C为钝角,同样方法得QN-PM=BC

若△ABC中∠B为钝角,同样方法得PM-QN=BC