Chứng minh tam giác ABD=tam giác IBD biết ABC vuông tại A, BD phân giác góc B

a/ Xét \(\Delta ABD;\Delta IBD\) có :

\(\left\{{}\begin{matrix}\widehat{BAC}=\widehat{BID}=90^0\\BDchung\\\widehat{B1}=\widehat{B2}\end{matrix}\right.\)

\(\Leftrightarrow\Delta ABD=\Delta IBD\left(ch-gn\right)\)

b/ \(\Delta ABD=\Delta IBD\left(cmt\right)\)

\(\Leftrightarrow AB=IB\)

Gọi giao điểm của BD và AI là H

Xét \(\Delta ABH;\Delta IBH\) có :

\(\left\{{}\begin{matrix}AB=IB\\\widehat{B1}=\widehat{B2}\\BHchung\end{matrix}\right.\)

\(\Leftrightarrow\Delta ABH=\Delta IBH\left(c-g-c\right)\)

\(\Leftrightarrow\widehat{AHB}=\widehat{IHB}\)

Mà \(\widehat{AHB}+\widehat{IHB}=180^0\)

\(\Leftrightarrow\widehat{AHB}=\widehat{IHB}=\dfrac{180^0}{2}=90^0\)

\(\Leftrightarrow BD\perp AI\left(đpcm\right)\)

c/ \(\Delta ABD=\Delta IBD\left(cmt\right)\)

\(\Leftrightarrow AD=ID\)

Xét \(\Delta AKD;\Delta ICD\) có :

\(\left\{{}\begin{matrix}\widehat{DAK}=\widehat{DIC}=90^0\\AD=ID\\\widehat{ADK}=\stackrel\frown{IDC}\end{matrix}\right.\)

\(\Leftrightarrow\Delta AKD=\Delta ICD\left(g-c-g\right)\)

\(\Leftrightarrow DK=DC\)