2008 AMC 10B Problems/Problem 24

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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]

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AMC 10 Problems and Solutions
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