Question

1.Fie MNPQ un patrat cu A ∈ MN, astfel incat AN\AM=1\3 si B ∈ PN astfel incat BN\PN=3\4 iar AQ∩BM=T. Aratati ca a. AQ=BM b. AQ⊥BM

Answer 1

a) AN/AM=1/3=>AM=3AN=>AN=AM/3

MN=MA+AN=3AN+AN=4AN=4AM/3

BN/PN=3/4=>PN=4BN/3

MNPQ->patrat=>MN=PN=>4AM/3=4BN/3|:4/3=>AM=BN=x

In ΔAMQ,m(∡AMQ)=90°=>(T.P.):AQ²=AM²+MQ²=AM²+MN²=AM²+(4AM/3)²=x²+16x²/9=(9x²+16x²)/9=25x²/9=>AQ=±√(25x²/9)=5x/3

AQ>0=>AQ=5x/3 (1)

In ΔMNB,m(∡MNB)=90°=>(T.P.):BM²=MN²+NB²=NP²+NB²=(4BN/3)²+BN²=x²+16x²/9=(9x²+16x²)/9=25x²/9=>BM=±√(25x²/9)=5x/3

BM>0=>BM=5x/3 (2)

Cum (1)=(2)=>AQ=BM

b) In ΔMNB,m(∡MNB)=90°=>sin(∡NBM)=MN/NB=4x/3/5x/3=4/5 (*)

In ΔAMQ,m(∡AMQ)=90°=>sin(∡QAM)=MQ/AQ4x/3/5x/3=4/5 (**)

Cum (*)=(**)=>m(∡NBM)=m(∡QAM)=α

m(∡NBM)+m(∡BMN)+m(∡MNB)=180°<=>α+90°+m(∡BMN)=180°=>m(∡BMN)=90-α (3)

m(∡AMQ)+m(∡MQA)+m(∡QAM)=180°<=>m(∡MQA)+90°+α=180°=>m(∡MQA)=90°-α (4)

Cum (3)=(4)=>m(∡MQA)=m(∡BMN)=β=90°-α

m(∡MAT)=m(∡MAQ)=α

m(∡NAT)=180°-m(∡MAT)=180°-α

m(∡ANB)=m(∡MNB)=90°

m(∡NBT)=m(∡NBM)=α

In patrulaterul NATB, m(∡ANB)+m(∡NBT)+m(∡BTA)m(∡TAN)=360°<=>90°+α+m(∡BTA)+180-α=360°<=>270°+m(∡BTA)=360°=>m(∡BTA)=90°

m(∡BTA)=m(∡MTQ)=90°( unghiuri opuse la varf)

m(∡ATM)=m(∡BTQ)=γ( unghiuri opuse la varf)

m(∡BTA)+m(∡MTQ)+m(∡ATM)+m(∡BTQ)=360°(unghiuri in jurul unui punct)=>90°+90°+γ+γ=360°=>2γ=180°=>γ=90°=>m(∡BTA)=m(∡MTQ)=m(∡ATM)=m(∡BTQ)=90°=>AT⊥BM=>AQ⊥BM

Answer 2

Răspuns:

Explicație pas cu pas:

ai totul in figura