

Line 3: 
Line 3: 
   
 <math>\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95</math>   <math>\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95</math> 
− 
 
−  ==Solution==
 
−  ===Solution 1 ===
 
−  Draw the angle bisectors of the angles <math>ABC</math> and <math>BCD</math>. These two bisectors obviously intersect. Let their intersection be <math>P</math>.
 
−  We will now prove that <math>P</math> lies on the segment <math>AD</math>.
 
− 
 
−  Note that the triangles <math>ABP</math> and <math>CBP</math> are congruent, as they share the side <math>BP</math>, and we have <math>AB=BC</math> and <math>\angle ABP = \angle CBP</math>.
 
− 
 
−  Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are congruent.
 
− 
 
−  Now we can compute their inner angles. <math>BP</math> is the bisector of the angle <math>ABC</math>, hence <math>\angle ABP = \angle CBP = 35^\circ</math>, and thus also <math>\angle CDP = 35^\circ</math>. (Faster Solution picks up here) <math>CP</math> is the bisector of the angle <math>BCD</math>, hence <math>\angle BCP = \angle DCP = 85^\circ</math>, and thus also <math>\angle BAP = 85^\circ</math>.
 
− 
 
−  It follows that <math>\angle APB = \angle BPC = \angle CPD = 180^\circ  35^\circ  85^\circ = 60^\circ</math>. Thus the angle <math>APD</math> has <math>180^\circ</math>, and hence <math>P</math> does indeed lie on <math>AD</math>. Then obviously <math>\angle BAD = \angle BAP = \boxed{ 85^\circ }</math>.
 
− 
 
−  <asy>
 
−  unitsize(1cm);
 
−  defaultpen(.8);
 
−  real a=4;
 
−  pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);
 
−  draw(ABCDcycle);
 
−  pair P1=B+3*a*dir(145), P2=C+3*a*dir(205);
 
−  pair P=intersectionpoint(BP1,CP2);
 
−  draw(BPC);
 
−  label("$A$",A,SW);
 
−  label("$B$",B,SE);
 
−  label("$C$",C,NE);
 
−  label("$D$",D,N);
 
−  label("$P$",P,W);
 
− 
 
−  label("$35^\circ$",B + dir(18017.5));
 
−  label("$35^\circ$",B + dir(1803517.5));
 
− 
 
−  label("$85^\circ$",C + .5*dir(120+42.5));
 
−  label("$85^\circ$",C + .5*dir(120+85+42.5));
 
−  </asy>
 
− 
 
−  Faster Solution: Because we now know three angles, we can subtract to get <math>360  35  85  85  35  35</math>, or <math>\boxed{85}</math>.
 
− 
 
− 
 
−  Even Faster Solution: Above, we proved that P falls on line AD, and also <math>\triangle ABP = \triangle CBP</math>, by <math>SAS</math>, hence we have <math>\angle BCP=\angle BAP</math>, which is the angle bisector of <math>\angle BCD</math> which is <math>\dfrac{170}{2}=85</math>. Hence we have <math>\angle BCP=\angle BAP=\angle BAD= 85^\circ</math>
 
− 
 
−  === Solution 2 ===
 
−  Draw the diagonals <math>\overline{BD}</math> and <math>\overline{AC}</math>, and suppose that they intersect at <math>E</math>. Then, <math>\triangle ABC</math> and <math>\triangle BCD</math> are both isosceles, so by anglechasing, we find that <math>\angle BAC = 55^{\circ}</math>, <math>\angle CBD = 5^{\circ}</math>, and <math>\angle BEA = 180  \angle EBA  \angle BAE = 60^{\circ}</math>. Draw <math>E'</math> such that <math>EE'B = 60^{\circ}</math> and so that <math>E'</math> is on <math>\overline{AE}</math>, and draw <math>E''</math> such that <math>\angle EE''C = 60^{\circ}</math> and <math>E''</math> is on <math>\overline{DE}</math>. It follows that <math>\triangle BEE'</math> and <math>\triangle CEE''</math> are both equilateral. Also, it is easy to see that <math>\triangle BEC \cong \triangle DE''C</math> and <math>\triangle BCE \cong \triangle BAE'</math> by construction, so that <math>DE'' = BE = EE'</math> and <math>EE'' = CE = E'A</math>. Thus, <math>AE = AE' + E'E = EE'' + DE'' = DE</math>, so <math>\triangle ADE</math> is isosceles. Since <math>\angle AED = 120^{\circ}</math>, then <math>\angle DAC = \frac{180  120}{2} = 30^{\circ}</math>, and <math>\angle BAD = 30 + 55 = 85^{\circ}</math>.
 
−  <asy>
 
−  import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=0.237,xmax=2.492,ymin=0.16,ymax=1.947;
 
−  pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0);
 
−  filldraw(arc((1,0),0.141,115,175)(1,0)cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)(0.658,0.94)cycle,evefev,qqwuqq); draw((0,0)(1,0)); draw((1,0)(0.658,0.94)); draw((0.658,0.94)(0.158,1.806)); draw((0.158,1.806)(0,0)); draw((0,0)(0.658,0.94)); draw((0.158,1.806)(1,0)); draw((0.058,0.082)(1,0)); draw((0.558,0.948)(0.658,0.94));
 
−  dot((0,0),ds); label("$A$",(0.096,0.005),NE*lsf); dot((1,0),ds); label("$B$",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label("$C$",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label("$D$",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label("$E$",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label("$E'$",(0.1,0.23),NE*lsf); label("$60^\circ$",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label("$E''$",(0.423,0.957),NE*lsf); label("$60^\circ$",(0.761,0.886),NE*lsf,qqwuqq);
 
−  clip((xmin,ymin)(xmin,ymax)(xmax,ymax)(xmax,ymin)cycle);
 
−  </asy>
 
− 
 
−  ===Solution 3 ===
 
−  Again, draw the diagonals <math>\overline{BD}</math> and <math>\overline{AC}</math>, and suppose that they intersect at <math>E</math>. We find by angle chasing the same way as in solution 2 that <math>m\angle ABE = 65^\circ</math> and <math>m\angle DCE = 115^\circ</math>. Applying the Law of Sines to <math>\triangle AEB</math> and <math>\triangle EDC</math>, it follows that <math>DE = \frac{2\sin 115^\circ}{\sin \angle DEC} = \frac{2\sin 65^\circ}{\sin \angle AEB} = EA</math>, so <math>\triangle AED</math> is isosceles. We finish as we did in solution 2.
 
− 
 
−  <asy>
 
−  unitsize(1cm);
 
−  defaultpen(.8);
 
−  real a=4;
 
−  pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);
 
−  draw(ABCDcycle);
 
−  pair P=intersectionpoint(BD,CA);
 
−  draw(AC); draw(BD);
 
−  label("$A$",A,SW);
 
−  label("$B$",B,SE);
 
−  label("$C$",C,NE);
 
−  label("$D$",D,N);
 
−  label("$E$",P,W);
 
−  </asy>
 
− 
 
−  === Solution 4 ===
 
−  Start off with the same diagram as solution 1. Now draw <math>\overline{CA}</math> which creates isosceles <math>\triangle CAB</math>. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that it's is <math>\boxed{85}.</math>
 
   
 ==Solution 5==   ==Solution 5== 