Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. Theo tc 2 tt cắt nhau: \(AC=AM;BM=BD\)
\(\Rightarrow AC+BD=AM+BM=AB\)
b. \(\left\{{}\begin{matrix}\widehat{AMO}=\widehat{ACO}=90^0\\AC=AM\\AO.chung\end{matrix}\right.\Rightarrow\Delta AOC=\Delta AOM \)
\(\Rightarrow\widehat{COA}=\widehat{AOM}=\dfrac{1}{2}\widehat{COM}\)
\(\left\{{}\begin{matrix}\widehat{ODB}=\widehat{OMB}=90^0\\BD=MB\\OB.chung\end{matrix}\right.\Rightarrow\Delta OBD=\Delta OBM\\ \Rightarrow\widehat{DOB}=\widehat{BOM}=\dfrac{1}{2}\widehat{DOM}\)
\(\Rightarrow\widehat{AOB}=\widehat{AOM}+\widehat{BOM}=\dfrac{1}{2}\left(\widehat{COM}+\widehat{DOM}\right)=\dfrac{1}{2}\cdot180^0=90^0\\ \Rightarrow\Delta OAB\text{ vuông tại O}\)
c. Áp dụng HTL: \(AM\cdot MB=OM^2=R^2\)
Mà \(CD=2R;AM=AC;BM=BD\)
Vậy \(AC\cdot BD=AM\cdot BM=R^2=\left(\dfrac{CD}{2}\right)^2=\dfrac{CD^2}{4}\)
A D B C O
ta có \(\frac{1}{AO^2}=\frac{1}{AB^2}+\frac{1}{AD^2}\Rightarrow AD=\sqrt{117}cm\)
ta có \(\widehat{ADB}=\widehat{ACD}\text{ (do cùng phụ với góc }\widehat{CDB}\text{)}\) nên \(\Delta ADB~\Delta DAC\left(g.g\right)\Rightarrow\frac{DA}{DC}=\frac{AB}{AD}\Rightarrow DC=\frac{DA^2}{AB}=\frac{9\sqrt{13}}{2}\)
ta có \(S_{ABCD}=\frac{1}{2}AD\left(AB+CD\right)=\frac{507}{4}cm^2\)
Lời giải:
a. Vì $AM$ là đường kính nên $\widehat{ABM}=90^0$ (góc nt chắn nửa đường tròn)
$\Rightarrow BM\perp AB$
Mà $CH\perp AB$ nên $BM\parallel CH(1)$
Tương tự: $\widehat{ACM}=90^0$ nên $AC\perp CM$
Mà $AC\perp BH$ nên $CM\parallel BH(2)$
Từ $(1); (2)$ suy ra $BHCM$ là hbh (tứ giác có 2 cặp cạnh đối song song)
b.
$\widehat{BAN}=90^0-\widehat{ABD}=90^0-\widehat{ABC}$
$=90^0-\widehat{AMC}$ (góc nt cùng chắn cung AC)
$=\widehat{MAC}$ (đpcm)
Vì $\widehat{BAN}=\widehat{MAC}$
$\Rightarrow \widehat{BAN}+\widehat{NAM}=\widehat{MAC}+\widehat{NAM}$
$\Leftrightarrow \widehat{BAM}=\widehat{CAN}$
$\Leftrightarrow \frac{1}{2}\text{sđc(BM)}=\frac{1}{2}\text{sđc(CN)}$
$\Leftrightarrow \widehat{BCM}=\widehat{CBN}(*)$
Lại có:
$\widehat{ANM}=90^0$ (góc nt chắn nửa đường tròn)
$\Rightarrow AN\perp MN$
Mà $AN\perp BC\Rightarrow MN\parallel BC$
$\Rightarrow BNMC$ là hình thang $(**)$
Từ $(*); (**)$ suy ra $BNMC$ là htc.
88888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888898888
??????????????????