Vì ΔAHB ~ ΔBCD (theo a)

=> \(\dfrac{HB}{CD}=\dfrac{AH}{BC}\Leftrightarrow\dfrac{2HI}{2MC}=\dfrac{AH}{BC}\Leftrightarrow\dfrac{HI}{MC}=\dfrac{AH}{BC}\)

Xét ΔAHI và ΔBCM có

\(\widehat{H}=\widehat{C}\left(=90^o\right)\)

\(\dfrac{HI}{MC}=\dfrac{AH}{BC}\left(cmt\right)\)

=> ΔAHI ~ ΔBCM (c-g-c)

=> \(\widehat{A1}=\widehat{B1}\) (2 góc tương ứng )

vì ABCD là hcn

=> AD// BC

=> \(\widehat{ADB}=\widehat{DBC}\left(SLT\right)\)

Xét ΔADH và ΔDBC có

\(\widehat{H}=\widehat{C}\left(=90^o\right)\)

\(\widehat{ADB}=\widehat{DBC}\left(cmt\right)\)

=> ΔADH ~ ΔDBC (g-g)

=> \(\widehat{DAH}=\widehat{BDC}\) (2 góc tương ứng) (1)

ta thấy

\(\widehat{AND}=\widehat{A1}+\widehat{E}\) ( \(\widehat{AND}\) là góc ngoài của ΔAEN)

\(\widehat{BMD}=\widehat{B1}+\widehat{C}\)( \(\widehat{BMD}\)là góc ngoài của Δ BMC)

Mà \(\widehat{B1}=\widehat{A1}\left(ctm\right)\)

\(\widehat{E}=\widehat{C}\left(=90^o\right)\)

=> \(\widehat{AND}=\widehat{BMD}\) (2)

Từ (1) và (2)

=> ΔAND ~ Δ DMB (g-g)

=> \(\dfrac{AN}{DM}=\dfrac{AD}{BD}\)

=> AN.BD=AD.DM (ĐPCM)

1 + 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

M = ( 71 + x ) - ( - 24 - x ) + ( - 35 - x )

   = 71 + x + 24 + x - 35 - x

  = ( x + x - x ) + ( 71 + 24 - 35 )

  = x + 60

xét hiệu

\(\dfrac{a^2+b^2+c^2}{3}-\dfrac{\left(a+b+c\right)^2}{9}\ge0\)

<=> \(\dfrac{3\left(a^2+b^2+c^2\right)}{9}-\dfrac{\left(a+b+c\right)^2}{9}\ge0\)

<=>\(3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2ac-2bc\ge0\)

<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)

<=> \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\)

<=> (a-b)² +(b-c)² +(c-a)² ≥ 0 (luôn đúng)

=> đpcm)