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

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 equal, 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 equal.

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]$

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2008 AMC 10B Problems/Problem 24

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