2008 AMC 10B Problems/Problem 24

Problem

Quadrilateral $ABCD$ has $AB = BC = CD$ , angle $ABC = 70$ and angle $BCD = 170$ . What is the measure of angle $BAD$ ?

$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$

Solution

Solution 1

Draw the angle bisectors of the angles $ABC$ and $BCD$ . These two bisectors obviously intersect. Let their intersection be $P$ . We will now prove that $P$ lies on the segment $AD$ .

Note that the triangles $ABP$ and $CBP$ are congruent, as they share the side $BP$ , and we have $AB=BC$ and $\angle ABP = \angle CBP$ .

Also note that for similar reasons the triangles $CBP$ and $CDP$ are congruent.

Now we can compute their inner angles. $BP$ is the bisector of the angle $ABC$ , hence $\angle ABP = \angle CBP = 35^\circ$ , and thus also $\angle CDP = 35^\circ$ . $CP$ is the bisector of the angle $BCD$ , hence $\angle BCP = \angle DCP = 85^\circ$ , and thus also $\angle BAP = 85^\circ$ .

It follows that $\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ$ . Thus the angle $APB$ has $180^\circ$ , and hence $P$ does indeed lie on $AD$ . Then obviously $\angle BAD = \angle BAP = \boxed{ 85^\circ }$ .

$[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,N); label("$P$",P,W); label("$35^\circ$",B + dir(180-17.5)); label("$35^\circ$",B + dir(180-35-17.5)); label("$85^\circ$",C + .5*dir(120+42.5)); label("$85^\circ$",C + .5*dir(120+85+42.5)); [/asy]$

Solution 2

Draw the diagonals $\overline{BD}$ and $\overline{AC}$ , and suppose that they intersect at $E$ . Then, $\triangle ABC$ and $\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\angle BAC = 55^{\circ}$ , $\angle CBD = 5^{\circ}$ , and $\angle BEA = 180 - \angle EBA - \angle BAE = 60^{\circ}$ . Draw $E'$ such that $EE'B = 60^{\circ}$ and so that $E'$ is on $\overline{AE}$ , and draw $E''$ such that $\angle EE''C = 60^{\circ}$ and $E''$ is on $\overline{DE}$ . It follows that $\triangle BEE'$ and $\triangle CEE''$ are both equilateral. Also, it is easy to see that $\triangle BEC \cong \triangle DE''C$ and $\triangle BCE \cong \triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$ . Thus, $AE = AE' + E'E = EE'' + DE'' = DE$ , so $\triangle ADE$ is isosceles. Since $\angle AED = 120^{\circ}$ , then $\angle DAC = \frac{180 - 120}{2} = 30^{\circ}$ , and $\angle BAD = 30 + 55 = 85^{\circ}$ . $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label("$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 $\overline{BD}$ and $\overline{AC}$ , and suppose that they intersect at $E$ . We find by angle chasing the same way as in solution 2 that $m\angle ABE = 65^\circ$ and $m\angle DCE = 115^\circ$ . Applying the Law of Sines to $\triangle AEB$ and $\triangle EDC$ , it follows that $DE = \frac{2\sin 115^\circ}{\sin \angle DEC} = \frac{2\sin 65^\circ}{\sin \angle AEB} = EA$ , so $\triangle AED$ is isosceles. We finish as we did in solution 2.

$[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P=intersectionpoint(B--D,C--A); draw(A--C); draw(B--D); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,N); label("$E$",P,W); [/asy]$

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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