Click vào số 0 bên trai cùng để vote cho mình nếu lời giải đúng nhé :DTa ki hieu d(I,PQ) la khoang cach tu I den PQ. Goi E′ la diem tren AC sao cho DM la phan giac cua BDE. Khi do d(M,AC)=d(M,AB)=d(M,DE′)⇒E′Mlaphangiaccua∠DE′CDo vay $ \angle DME' =180^0-\frac{1}{2}(\angle BDE' +\angle DE'C) =180^0-\frac{1}{2}(360^0-\angle B-\angle C)\\=180^0-180^0 +\angle B =\angle B=\angle MDE .DovayE\equiv E'.Dieudosuyrab)BDlaphangiaccua\angle BDEa)\triangle DBM \sim \triangle DME \sim \triangle MCE \Rightarrow \frac{BD}{MB}=\frac{MC}{CE} \Rightarrow BD.CE =MB.MC =\frac{BC^2}{4} c)HaMB' \perp AB, MC'\perp AC, MA'\perp DE,voidieukien\triangle ABCdeu,tacoAD+DE+AE=AD+DA′+A′E+AE=AD+DB′+EC′+AE=AB′+AC′=2.AB′=2AM2AB=23a2/4a=32a$
Click vào số 0 bên trai cùng để vote cho mình nếu lời giải đúng nhé :DTa ki hieu d(I,PQ) la khoang cach tu I den PQ. Goi E′ la diem tren AC sao cho DM la phan giac cua BDE. Khi do d(M,AC)=d(M,AB)=d(M,DE′)⇒E′Mlaphangiaccua∠DE′CDo vay $ \angle DME' =180^0-\frac{1}{2}(\angle BDE' +\angle DE'C) =180^0-\frac{1}{2}(360^0-\angle B-\angle C) =180^0-180^0 +\angle B =\angle B=\angle MDE .DovayE\equiv E'.Dieudosuyrab)BDlaphangiaccua\angle BDEa)\triangle DBM \sim \triangle DME \sim \triangle MCE \Rightarrow \frac{BD}{MB}=\frac{MC}{CE} \Rightarrow BD.CE =MB.MC =\frac{BC^2}{4} c)HaMB' \perp AB, MC'\perp AC, MA'\perp DE,voidieukien\triangle ABCdeu,tacoAD+DE+AE=AD+DA′+A′E+AE=AD+DB′+EC′+AE=AB′+AC′=2.AB′=2AM2AB=23a2/4a=32a$