Question

Cho hình bình hành \(ABCD\). \(M\) là trung điểm \(BC\). Trên \(AB\) lấy điểm \(N\) sao cho \(\overrightarrow{BN}=\dfrac{1}{3}\overrightarrow{BA}\), \(CN\) giao \(DM\) tại \(K\).

a) Hãy biểu diễn \(\overrightarrow{MN}\) theo \(\overrightarrow{DB},\overrightarrow{DC}\)

b) Tính \(\dfrac{KN}{KC}\)

c) Dựng điểm \(X\) thỏa mãn: \(\overrightarrow{AX}+2\overrightarrow{BX}=4\overrightarrow{DX}-\overrightarrow{CX}\)

d) \(\overrightarrow{AB}=\left(x+1\right)\overrightarrow{AP}\)    

    \(\overrightarrow{AD}=\left(3x-1\right)\overrightarrow{AQ}\)

CMR: \(PQ\) đi qua một điểm cố định

Answer 1

d.

Gọi H là điểm sao cho \(\overrightarrow{AH}=\dfrac{3}{4}\overrightarrow{AB}-\dfrac{1}{4}\overrightarrow{AD}\)

Do ABCD cố định \(\Rightarrow H\) cố định

Ta có:

\(\overrightarrow{AB}=\left(x+1\right)\overrightarrow{AP}=\left(x+1\right)\left(\overrightarrow{AH}+\overrightarrow{HP}\right)=\left(x+1\right)\overrightarrow{AH}+\left(x+1\right)\overrightarrow{HP}\)

\(=\left(x+1\right)\left(\dfrac{3}{4}\overrightarrow{AB}-\dfrac{1}{4}\overrightarrow{AD}\right)+\left(x+1\right)\overrightarrow{HP}\)

\(\Rightarrow\left(x+1\right)\overrightarrow{HP}=\overrightarrow{AB}-\left(x+1\right)\left(\dfrac{3}{4}\overrightarrow{AB}-\dfrac{1}{4}\overrightarrow{AD}\right)=-\dfrac{3x-1}{4}\overrightarrow{AB}+\dfrac{x+1}{4}\overrightarrow{AD}\)

\(\overrightarrow{AD}=\left(3x-1\right)\overrightarrow{AQ}=\left(3x-1\right)\left(\overrightarrow{AH}+\overrightarrow{HQ}\right)=\left(3x-1\right)\overrightarrow{AH}+\left(3x-1\right)\overrightarrow{HQ}\)

\(=\left(3x-1\right)\left(\dfrac{3}{4}\overrightarrow{AB}-\dfrac{1}{4}\overrightarrow{AD}\right)+\left(3x-1\right)\overrightarrow{HQ}\)

\(\Rightarrow\left(3x-1\right)\overrightarrow{HQ}=\overrightarrow{AD}-\left(3x-1\right)\left(\dfrac{3}{4}\overrightarrow{AB}-\dfrac{1}{4}\overrightarrow{AD}\right)=-\dfrac{3}{4}\left(3x-1\right)\overrightarrow{AB}+3.\dfrac{x+1}{4}\overrightarrow{AD}\)

\(=3\left[-\dfrac{3x-1}{4}\overrightarrow{AB}+\dfrac{x+1}{4}\overrightarrow{AD}\right]\)

\(\Rightarrow\left(3x-1\right)\overrightarrow{HQ}=3.\left(x+1\right)\overrightarrow{HP}\)

\(\Rightarrow H,P,Q\) thẳng hàng hay PQ luôn đi qua H cố định

Answer 2

a.

\(\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{BA}=\dfrac{1}{2}\overrightarrow{DA}-\dfrac{1}{3}\overrightarrow{DC}\)

\(=\dfrac{1}{2}\left(\overrightarrow{DA}+\overrightarrow{DC}\right)-\dfrac{1}{2}\overrightarrow{DC}-\dfrac{1}{3}\overrightarrow{DC}=\dfrac{1}{2}\overrightarrow{DB}-\dfrac{5}{6}\overrightarrow{DC}\)

b.

\(\overrightarrow{CN}=\overrightarrow{CB}+\overrightarrow{BN}=\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{BA}=\overrightarrow{CB}-\dfrac{1}{3}\overrightarrow{DC}\)

Đặt \(\overrightarrow{CK}=x.\overrightarrow{CN}\)

\(\overrightarrow{DK}=\overrightarrow{DC}+\overrightarrow{CK}=\overrightarrow{DC}+x\overrightarrow{CN}=\overrightarrow{DC}+x.\overrightarrow{CB}-\dfrac{x}{3}\overrightarrow{DC}=x.\overrightarrow{CB}+\left(1-\dfrac{x}{3}\right)\overrightarrow{DC}\)

\(\overrightarrow{DM}=\overrightarrow{DC}+\overrightarrow{CM}=\dfrac{1}{2}\overrightarrow{CB}+\overrightarrow{DC}\)

D, K, M thẳng hàng \(\Rightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{1-\dfrac{x}{3}}{1}\Rightarrow x=\dfrac{3}{7}\)

\(\Rightarrow CK=\dfrac{3}{7}CN\Rightarrow7KC=3CN=3\left(KC+KN\right)\)

\(\Rightarrow4KC=3KN\Rightarrow\dfrac{KN}{KC}=\dfrac{4}{3}\)

Answer 3

c.

\(\overrightarrow{AX}+2\overrightarrow{BX}=4\overrightarrow{DX}-\overrightarrow{CX}\)

\(\Leftrightarrow\left(\overrightarrow{AX}-\overrightarrow{DX}\right)+2\left(\overrightarrow{BX}-\overrightarrow{DX}\right)=\overrightarrow{DX}-\overrightarrow{CX}\)

\(\Leftrightarrow\left(\overrightarrow{AX}+\overrightarrow{XD}\right)+2\left(\overrightarrow{BX}+\overrightarrow{XD}\right)=\overrightarrow{DX}+\overrightarrow{XC}\)

\(\Leftrightarrow\overrightarrow{AD}+2\overrightarrow{BD}=\overrightarrow{DC}\)

\(\Leftrightarrow\overrightarrow{AD}-2\left(\overrightarrow{DA}+\overrightarrow{DC}\right)=\overrightarrow{DC}\)

\(\Leftrightarrow3\overrightarrow{AD}=3\overrightarrow{DC}\)

\(\Leftrightarrow\overrightarrow{AD}=\overrightarrow{DC}\) (vô lý)

Ko có điểm X thỏa mãn yêu cầu.