Question

Buna.Imi puteti rezolva o problema de geometrie?

Answer 1

solutie:a) ABCD = patrat ⇒  AB = BC = CD = AD =4 cm .ΔABN = Δ echilateral⇒AN=AB=BN=4cm . PNADCB=NA+AD+DC+BC+BN⇒P NADCB=5·4⇒P NADCB=20 cm   b)ΔANB =Δechilateral⇒m(NAB) = m(ABN) = m(ANB) = 60 GRADE.  m(NAM)=m(NAB) + m(BAM)⇒m(NAM)=60+30⇒m(NAM)=90 grade⇒NA perpendicular AM ; fie {P}=AB intersectat cu MN ;ΔMAN≡ΔMBN⇒AMN≡BMN, P∈MN⇒AMP≡BMN(1) Cum AM = BM (ip) ,AMP≡BMN(cf 1),MP=PM(latura comuna)⇒ΔAMP≡ΔBMP⇒APM≡BPM. P∈AB⇒ A,P,B = pc coliniare⇒m(APB)=180. Cum m (APB)=180 ,m (APB)=m(APM)+m(BPM) si  APM≡BPM⇒m(APM)=m(BPM)=90. Cum m(APB)=90⇒NM perpendicular pe AB.