Question

Answer 1

a,Xét Δ NMB và Δ NDC có

MB // DC => Δ NMB ∼ Δ NDC 

Xét Δ AKD và ΔCKN có

\(\widehat{AKD}=\widehat{NKC}\left(\text{đ}\text{ối}\text{đ}\text{ỉnh}\right)\)

\(\widehat{DAK}=\widehat{NCK}\) (soletrong)

=> Δ AKD ∼ ΔCKN  (g.g)

b, Xét ΔAMK và ΔCDK có

\(\widehat{AKM}=\widehat{DKC}\) (đối đỉnh)

\(\widehat{KAM}=\widehat{KCD}\) (soletrong)

=> ΔAMK ∼ ΔCDK (g.g)

=> \(\dfrac{KM}{KD}=\dfrac{AK}{CK}\) (1)

Ta có Δ AKD ∼ ΔCKN

=> \(\dfrac{AK}{CK}=\dfrac{KD}{KN}\)(2)

Từ (1),(2) => \(\dfrac{KM}{KD}=\dfrac{KD}{KN}\) => KD² = KM . KN

c, Ta có Δ NMB ∼ Δ NDC 

=> \(\dfrac{MB}{DC}=\dfrac{NB}{NC}=\dfrac{NM}{ND}=\dfrac{2}{5}\) 

=> Tỉ số đồng dạng = \(\dfrac{2}{5}\)

=> DC = 10

BC = 15 - 6 = 9

=> S_ABCD = 10 . 9 = 90 (cm²)