Question

Answer 1

1: Ta có: \(\hat{ABP}=\hat{ABC}+\hat{CBP}=90^0+\hat{CBP}\)

\(\hat{MBC}=\hat{MBP}+\hat{CBP}=90^0+\hat{CBP}\)

Do đó: \(\hat{ABP}=\hat{BMC}\)

Xét ΔABP và ΔMBC có

BA=BM

\(\hat{ABP}=\hat{MBC}\)

BP=BC

Do đó; ΔABP=ΔMBC

Ta có: \(\hat{CBP}+\hat{C^{\prime}BP}=\hat{CBC^{\prime}}=90^0\)

\(\hat{C^{\prime}BM}+\hat{C^{\prime}BP}=\hat{MBP}=90^0\)

Do đó: \(\hat{CBP}=\hat{C^{\prime}BM}\)

Xét ΔCBP và ΔC'BM có

CB=C'B

\(\hat{CBP}=\hat{C^{\prime}BM}\)

BP=BM

Do đó: ΔCBP=ΔC'BM