Giải thích các bước giải:
a.Ta có :A D / / B C → B M A ^ = D A N ^ AD//BC\to\widehat{BMA}=\widehat{DAN} A D / / B C → B M A = D A N
A B / / D N → B A M ^ = A N D ^ AB//DN\to \widehat{BAM}=\widehat{AND} A B / / D N → B A M = A N D
→ Δ A B M ∼ Δ N D A ( g . g ) \to\Delta ABM\sim\Delta NDA(g.g) → Δ A B M ∼ Δ N D A ( g . g )
→ B M D A = A B D N → B M . D N = A B . A D \to\dfrac{BM}{DA}=\dfrac{AB}{DN}\to BM.DN=AB.AD → D A B M = D N A B → B M . D N = A B . A D
b.Ta có :A D / / B M → A P P M = P D P B AD//BM\to\dfrac{AP}{PM}=\dfrac{PD}{PB} A D / / B M → P M A P = P B P D
→ A P A P + P M = P D P D + P B \to\dfrac{AP}{AP+PM}=\dfrac{PD}{PD+PB} → A P + P M A P = P D + P B P D → A P A M = P D B D \to\dfrac{AP}{AM}=\dfrac{PD}{BD} → A M A P = B D P D
Lại có :
A B / / D N → P A P N = P B P D AB//DN\to\dfrac{PA}{PN}=\dfrac{PB}{PD} A B / / D N → P N P A = P D P B → P A P A + P N = P B P D + P B \to\dfrac{PA}{PA+PN}=\dfrac{PB}{PD+PB} → P A + P N P A = P D + P B P B
→ A P A N = P B B D \to\dfrac{AP}{AN}=\dfrac{PB}{BD} → A N A P = B D P B
→ A P A M + A P A N = P D B D + P B B D \to\dfrac{AP}{AM}+\dfrac{AP}{AN}=\dfrac{PD}{BD}+\dfrac{PB}{BD} → A M A P + A N A P = B D P D + B D P B
→ A P A M + A P A N = P D + P B B D \to\dfrac{AP}{AM}+\dfrac{AP}{AN}=\dfrac{PD+PB}{BD} → A M A P + A N A P = B D P D + P B
→ A P A M + A P A N = B D B D \to\dfrac{AP}{AM}+\dfrac{AP}{AN}=\dfrac{BD}{BD} → A M A P + A N A P = B D B D
→ A P A M + A P A N = 1 \to\dfrac{AP}{AM}+\dfrac{AP}{AN}=1 → A M A P + A N A P = 1
→ 1 A M + 1 A N = 1 A P \to\dfrac{1}{AM}+\dfrac1{AN}=\dfrac1{AP} → A M 1 + A N 1 = A P 1
