Question

trapezul dreptunghic ABCD are AD||BC,AD<BC,AB perpend BC si CD=AD+BC.Fie M∈(CD),a.i. DM=AD.

a)Dem.ca m(<AMB)=90 gr

b)Perpendiculara in M pe CD,intersecteaza latura (AB) in N.Aratati ca m(<DNC)=90 gr

Answer 1

Δ ADM = isoscel (AD = DM) ⇒ mas<DAM = mas<DMA = mas<C /2
(deoarece mas<D = 180* - mas<C )
Δ BCM  MC = BC  (DC = DM + BC) ⇒mas<MBC = mas<CMB = (180* - mas<C) /2 =
= 90* - mas<C/2  ⇒ mas< AMB = 180* - (mas<C /2 + 90* - mas<C /2) = 180* - 90*
mas<AMB = 90*
b)Δ MNC ≡ Δ BNC (MC = BC, NC = lat. comuna) ⇒ mas<MCN= mas<NCB = mas<C /2
Δ ADN ≡ΔMDN (AD = DM) ⇒ mas<ADN = mas<MDN = mas<D /2 = (180* - mas<C)/2 =90* - mas<C /2
ΔDNC  mas<DNC = 180*- (mas<C /2 + 90* - mas<C /2) = 90*