Question

已知如图,在等腰三角形ABCD中,AD平行与BC,AB=CD.AD=3,BC=9,tan∠ABC等于4/3 ,直线MN是

Answer 1

作AG⊥BC,垂足G,作DH⊥BC,垂足H,

GH=AD=3,BG=HC=(BC-GH)/2=(9-3)/2=3;

tan∠ABC=4/3=AG/BG,

AG=BG*4/3=3*4/3=4,

MN=AG=3;

若PD‖AB‖CF,

∠ECF=∠PDC,[内错角]

∠F=∠ABP,[内错角]

等腰三角形ABCD,

∠ABC=∠DCB,

直线MN是梯形的对称轴,PB=PC,

∠PBN=∠PCN,

∠DCP=∠DCB-∠PCN=∠ABC-∠PBN=∠ABP=∠F;[∠ABC=∠DCB,∠F=∠ABP]

∠CEF=180°-∠F-∠ECF=180°-∠DCP-∠PDC=∠DPC,

△EFC∽△PDC,[AAA]

延长DP与BC交于J,PD‖AB,PJ‖AB,

∠ADJ=∠DJC,[内错角]

∠DJC=∠ABC,[同位角]

所以∠ADJ=∠ABC,

tan∠ADJ=tan∠ABC=4/3,

MP/MD=4/3,

MP=MD*4/3=(AD/2)*4/3=(3/2)*4/3=2,

PN=MN-MP=4-2=2.