d) Chứng minh BM vuông góc DH
Giải thích
d) Vì ΔBHO=ΔCBMc−c−c nên BHO^=MBC^
Gọi I=BM∩HD
IBM^=1800=IBH^+HBC^+CBM^=IBH^+900+CBM^⇒IBH^+CBM^=900⇒IBH^+BHO^=900⇒BIH^=900
Suy ra BM⊥DH.
d) Vì ΔBHO=ΔCBMc−c−c nên BHO^=MBC^
Gọi I=BM∩HD
IBM^=1800=IBH^+HBC^+CBM^=IBH^+900+CBM^⇒IBH^+CBM^=900⇒IBH^+BHO^=900⇒BIH^=900
Suy ra BM⊥DH.