Question

Answer 1

a) Tứ giác ABCD có: AD//BC, AB//CD.

\(\Rightarrow\)ABCD là hình bình hành.

b) △BAE và △BDF có: \(\widehat{AEB}=\widehat{DFB}=90^0;\widehat{DBF}\) là góc chung.

\(\Rightarrow\)△BAE∼△BDF (g-g)

\(\Rightarrow\dfrac{BA}{BD}=\dfrac{BE}{BF}\Rightarrow BA.BF=BD.BE\left(1\right)\)

c) △AED và △BKD có: \(\widehat{AED}=\widehat{BKD}=90^0;\widehat{ADE}=\widehat{BDK}\)

\(\Rightarrow\)△AED∼△BKD (g-g)

\(\Rightarrow\dfrac{DE}{DK}=\dfrac{DA}{DB}\Rightarrow DE.DB=DA.DK\left(2\right)\)

Từ (1), (2) suy ra: \(BA.BF+DA.DK=BD.BE+DE.DB=BD\left(BE+DE\right)=BD.BD=BD^2\)