Question

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Answer 1

IPOTEZA ABC triunghi, AD bisectoatre, D∈BC

[AM]≡[MD], M∈AD

MP⊥AD, MP∩AC={P}

CONCLUZIE: PD||AB

DEMO: mas ∡BAD=mas∡DAP (AD bisectoare, din ipoteza) (1)

P∈mediatoareAD(ipopteza)⇒[PA]≡[PD]⇒ΔPADisoscel⇒

⇒mas∡DAP=mas∡PDA (2)

din(1) si(2)⇒mas∡DAP=mas∡PDA dar PD , AB drepte, AD secanta,

iar ∡DAP si ∡PDA alterne interne(3)

din (2) si (3)∡DAP si ∡PDA alterne interne congruente⇒PD||AB Q.E.D.

Answer 2

Ipoteza:

ABC- triunghi

AD-bisectoare, D∈BC

M∈AD, [MA] ≡ [MD]

MP ⊥ AD, P ∈ AC

Concluzia :

PD || AB

Demonstrația:

$\it AD-bisectoare \Rightarrow \widehat{DAC} \equiv \widehat{DAB}\ \ \ \ \ (1)$

$\it M\in AD, \ [MA] \equiv [MD],\ MP \perp AD, P \in AC \Rightarrow MP-mediatoare\ pentru\ [AD] \Rightarrow \\ \\ \Rightarrow PA=PD \Rightarrow \Delta PAD-isoscel \Rightarrow \widehat{PAD} \equiv\widehat{PDA} \Rightarrow \widehat{DAC} \equiv\widehat{PDA}\ \ \ \ \ \ (2) \\ \\ (1),\ (2) \Rightarrow \widehat{DAB} \equiv\widehat{PDA} \ (alterne\ interne\ pentru\ secanta\ AD) \Rightarrow PD||AB\ [q.e.d.]$