Đáp án:

a) $\widehat{ABC}=\widehat{ACB}$

b) $AM=KC$

c) $\widehat{MAI}=\widehat{APC}$

Giải thích các bước giải:

a)

Xét $\triangle AIB$ và $\triangle AIC$:

$IB=IC$ (gt)

$\widehat{AIB}=\widehat{AIC}\,\,\,(=90^o)$

$AI$: chung

$\to\triangle AIB=\triangle AIC$ (c.g.c)

$\to AB=AC$ (2 cạnh tương ứng)

$\to\widehat{ABI}=\widehat{ACI}$ (2 góc tương ứng)

Hay $\widehat{ABC}=\widehat{ACB}$

b)

Ta có: $BK=AB$ (gt), $AB=AC$ (cmt)

$\to BK=AC$

Lại có:

$\widehat{ABI}+\widehat{KBC}=180^o$ (kề bù)

$\widehat{ACI}+\widehat{MCA}=180^o$ (kề bù)

Mà $\widehat{ABC}=\widehat{ACB}$ (cmt)

$\to\widehat{KBC}=\widehat{MCA}$

Xét $\triangle KBC$ và $\triangle ACM$:

$BK=CA$ (cmt)

$\widehat{KBC}=\widehat{ACM}$ (cmt)

$BC=CM$ (gt)

$\to\triangle KBC=\triangle ACM$ (c.g.c)

$\to KC=AM$ (2 cạnh tương ứng)

c)

Ta có: $\widehat{MAI}+\widehat{AMI}=90^o$ (2 góc phụ nhau)

Hay $\widehat{MAI}+\widehat{CMA}=90^o$

$\widehat{IPC}+\widehat{ICP}=90^o$ (2 góc phụ nhau)

Hay $\widehat{APC}+\widehat{BCK}=90^o$

Lại có: $\triangle KBC=\triangle ACM$ (cmt)

$\to\widehat{BCK}=\widehat{CMA}$ (2 góc tương ứng)

$\to\widehat{MAI}=\widehat{APC}$