Answer:
Number Thirteen:
[tex]because:\ \bar{AB}~\bar{CB},[/tex]
[tex]Then\ \bar{BD}\ is\ a\ median\ of\ \vec{AC}[/tex]
[tex]So:\bar{AD}\begin{matrix}n\\C\\\end{matrix}\bar{CD}[/tex]
[tex]Then\ \bar{BD}\cong~\begin{matrix}\boxdot\\BD\\\end{matrix}[/tex]
[tex]S_B\triangle ABD=\triangle CBD(SSS)[/tex]
[tex]cinvestigate\ provement\ 1[/tex]
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Number Twelve:
[tex]Solution: \{As\ shown\ in\ the\ figure\}[/tex]
[tex]because\ C\ is\ midpoint[/tex]
[tex]of\ \bar{AB}\ and\ \bar{AD}[/tex]
[tex]get\ that\ AC=CE.\ \ \ \ \ BC=CD[/tex]
[tex]and < ACB= < DCE[/tex]
[tex]So\ \triangle ABC\sim \triangle BDC[/tex]
[tex]\left\{\left\{\begin{matrix}AC=EC\\BC=DC\\ < ACB= < DCE\\\end{matrix}\right.\Rightarrow \triangle ABC \sim \triangle EDC\right\}[/tex]
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Number Eleven:
Answer:
[tex]\therefore < ADB=\triangle CDB,Proved[/tex]
Explanation:
[tex]qu < ADB\ and \triangle CBD[/tex]
[tex]< ADB=(CDB=90\textdegree (given)[/tex]
[tex]< A= < C \ \ \ \ \ (given)[/tex]
[tex]BD=BD(common)[/tex]
[tex]\therefore \triangle ADB\ \cong \triangle CDB,\ criteria\ (AAS)[/tex]
[tex]\therefore < ADB=\triangle CDB,\ Proved[/tex]
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Number Eighteen:
[tex]4s.\ Since\ \bar{AB}\bot\bar{BC},\ \bar{DE}\bot\bar{EF},\ thus\ < B\ and\ < E\ are\ right\ angles[/tex]
[tex]the\ meaning\ of\ vertical[/tex]
[tex]5.\ Since\ \bar{BC}/\bar{/FE},\ thus\ < BCA\ \cong < EFD[/tex]
[tex]if\ tho\ lines\ are\ parallel,\ affernate\ interior\ angles\ are\ congtuent.[/tex]
[tex]8.\ \triangle ABC\subseteq \ \triangle DOF\ \ \ \ \ AAS\ postulate[/tex]
I hope this helps you
:)
