# 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 $\angle CDP = 35^\circ$. (Faster Solution picks up here) $CP$ 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 $APD$ has $180^\circ$, and hence $P$ does indeed lie on $AD$. Then obviously $\angle BAD = \angle BAP = \boxed{ 85^\circ }$.

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

Faster Solution: Because we now know three angles, we can subtract to get $360 - 35 - 85 - 85 - 35 - 35$, or $\boxed{85}$.

Even Faster Solution: Above, we proved that P falls on line AD, and also $\triangle ABP = \triangle CBP$, by $SAS$, hence we have $\angle BCP=\angle BAP$, which is the angle bisector of $\angle BCD$ which is $\dfrac{170}{2}=85$. Hence we have $\angle BCP=\angle BAP=\angle BAD= 85^\circ$

### 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.

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

### Solution 4

Start off with the same diagram as solution 1. Now draw $\overline{CA}$ which creates isosceles $\triangle CAB$. 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 $\boxed{85}.$

## Solution 5

This solution requires the use of cyclic quadrilateral properties but could be a bit time-consuming during the contest. To start off, draw a diagram like in solution one and label the points. Now draw the $\overline{AC}$ and $\overline{BD}$ and call this intersection point $Y$. Note that triangle $BCD$ is an isosceles triangle so angles $CDB$ and $CBD$ are each $5$ degrees. Since $AB$ equals $BC$, angle $BAC$ had to equal $55$ degrees, thus making angle $AYB$ equal to $60$ degrees. We can also find out that angle CYB equals $120$ degrees. Extend point $C$ such that it lies on the same level of segment $AB$. Call this point $E$. Since angle $BEC$ plus angle $CYB$ equals $180$ degrees, quadrilateral $YCEB$ is a cyclic quadrilateral. Next, draw a line from point $Y$ to point $E$. Since angle $YBC$ and angle $YEC$ point to the same arc, angle $YEC$ is equal to $5 degrees$. Since $EBD$ is an isosceles triangle(based on angle properties) and $YAE$ is also an isosceles triangle, we can find that $YAD$ is also an isosceles triangle. Thus, each of the other angles is $\frac{180-120}{2}=30$ degrees. Finally, we have angle $BAD$ equals $30+55=\boxed{85}$ degrees.

## Solution 6

First, connect the diagonal $DB$, then, draw line $DE$ such that it is congruent to $DC$ and is parallel to $AB$. Because triangle $DCB$ is isosceles and angle $DCB$ is $170^\circ$, the angles $CDB$ and $CBD$ are both $\frac{180-170}{2} = 5^\circ$. Because angle $ABC$ is $70^\circ$, we get angle $ABD$ is $65^\circ$. Next, noticing parallel lines $AB$ and $DE$ and transversal $DB$, we see that angle $BDE$ is also $65^\circ$, and subtracting off angle $CDB$ gives that angle $EDC$ is $60^\circ$.

Now, because we drew $ED = DC$, triangle $DEC$ is equilateral. We can also conclude that $EC=DC=CB$ meaning that triangle $ECB$ is isosceles, and angles $CBE$ and $CEB$ are equal.

Finally, we can set up our equation. Denote angle $BAD$ as $x^\circ$. Then, because $ABED$ is a parallelogram, the angle $DEB$ is also $x^\circ$. Then, $CEB$ is $(x-60)^\circ$. Again because $ABED$ is a parallelogram, angle $ABE$ is $(180-x)^\circ$. Subtracting angle $ABC$ gives that angle $CBE$ equals $(110-x)^\circ$. Because angle $CBE$ equals angle $CEB$, we get $$x-60=110-x$$, solving into $x=\boxed{85^\circ}$.

$[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120), E=D+a*dir(0); draw(A--B--C--D--cycle); draw(E--C); draw(B--D); draw(B--E); draw(D--E); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,N); label("E",E,NE); label("60^\circ",C + .5*dir(360-65-115-55-30)); label("65^\circ",B + .75*dir(180-32.5)); label("x^\circ",A + .5*dir(42.5)); label("5^\circ",D + 2.5*dir(360-60-2.5)); label("60^\circ",D + .75*dir(360-30)); label("60^\circ",E + .5*dir(360-150)); label("5^\circ",B + 2.5*dir(180-65-2.5)); [/asy]$

Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles.

~Someonenumber011

## See also

 2008 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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