a) + Có: ABCD là hình thang cân (gt)
⇒ $\begin{cases} \text{AD = BC (tính chất)}\\ \text{$\widehat{ABD}$ = $\widehat{ABC}$ (t/c)}\\ \text{$\widehat{BCD}$ = $\widehat{ADC}$ (t/c)} \\ \text{AC = BD (t/c)} \end{cases}$
+ Xét ΔABD và ΔBAC có:
$\left.\begin{matrix} \text{Cạnh AB chung} \\ \text{$\widehat{BAD}$ = $\widehat{ABC}$ (cmt)} \\ \text{AD = BC (cmt)} \end{matrix}\right\}\text{⇒ ΔABD $\backsim$ ΔBAC (c - g - c)}$
⇒ $\widehat{ADB}$ = $\widehat{BCA}$ (2 góc tương ứng)
+ Xét ΔADC và ΔBCD có:
$\left.\begin{matrix} \text{Cạnh CD chung} \\ \text{$\widehat{ADC}$ = $\widehat{BCD}$ (cmt)} \\ \text{AD = BC (cmt)} \end{matrix}\right\}\text{⇒ ΔADC $\backsim$ ΔBCD (c - g - c)}$
⇒ $\widehat{DAC}$ = $\widehat{CBD}$ (2 góc tương ứng)
+ Xét ΔAOD và ΔBOC có:
$\left.\begin{matrix} \text{$\widehat{OAD}$ = $\widehat{OBC}$ (hay $\widehat{DAC}$ = $\widehat{CBD}$ - cmt)} \\ \text{AD = BC (cmt)} \\ \text{$\widehat{ADO}$ = $\widehat{BCO}$ (hay $\widehat{ADB}$ = $\widehat{BCA}$ - cmt)} \end{matrix}\right\}⇒ \text{ΔAOD $\backsim$ ΔBOC (g - c - g)}$
⇒ AO = BO = 4 cm (2 cạnh t/ứ)
OD = OC = 5 cm (2 cạnh t/ứ)
+ Có: OA + OC = AC (O ∈ AC)
T/số: 4 + 5 = AC
⇒ AC = 9 (cm)
mà AC = BD (cmt)
⇒ BD = AC = 9 cm
Bài 2:
Có: ABCD là hình thang cân (gt)
⇒ $\widehat{DAB}$ = $\widehat{ABC}$ (đ/n)
$\widehat{BCD}$ = $\widehat{ADC}$ = $70^o$(đ/n)
+ Có: $\widehat{DAB}$ + $\widehat{ABC}$ + $\widehat{BCD}$ + $\widehat{ADC}$ = $360^o$
T/số: $\widehat{DAB}$ + $\widehat{ABC}$ + $70^{o}$ $+^{}$ $70^{o}$ $=^{}$ $360^{o}$
$2\widehat{ABC}$ = $360^{o}$ $-^{}$ $70^{o}$ $-^{}$ $70^{o}$
$2\widehat{ABC}$ = $220^{o}$
⇒ $\widehat{DAB}$ = $2\widehat{ABC}$ = $110^{o}$
@tryphena
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