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 1

Note: 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. Create lines $\overline{AC}$ and $\overline{BD}$ . We can call their 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 $BAC$ equals $55$ degrees, thus making angle $AYB$ equal to $60$ degrees. We can also find out that angle $CYB$ equals $120$ degrees.

Extend $\overline{CD}$ and $\overline{AB}$ and let their intersection be $E$ . 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 $EYD$ 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.

~Minor edits by BakedPotato66

Solution 2

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$",C,SE); label("$D$",D,N); label("$E$",E,NE); label("$60^\circ$",C + .75*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

Solution 3(Using Trig.)

$[asy] unitsize(3 cm); pair A, B, C, D; A = (0,0); B = dir(85); C = B + dir(-25); D = C + dir(-35); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(((A + B)/2 + scale(0.02)*rotate(90)*(B - A))--((A + B)/2 + scale(0.02)*rotate(90)*(A - B))); draw(((B + C)/2 + scale(0.02)*rotate(90)*(C - B))--((B + C)/2 + scale(0.02)*rotate(90)*(B - C))); draw(((C + D)/2 + scale(0.02)*rotate(90)*(D - C))--((C + D)/2 + scale(0.02)*rotate(90)*(C - D))); dot("$A$", A, SW); dot("$B$", B, N); dot("$C$", C, NE); dot("$D$", D, SE); label("$I$", 6/7*C); [/asy]$

Let the unknown $\angle BAD$ be $x$ .

First, we draw diagonal $BD$ and $AC$ . $I$ is the intersection of the two diagonals. The diagonals each form two isosceles triangles, $\triangle BCD$ and $\triangle ABC$ .

Using this, we find: $\angle DBC = \angle CDB = 5^\circ$ and $\angle BAC = \angle BCA = 55^\circ$ . Expanding on this, we can fill in a couple more angles. $\angle ABD = 70^\circ - 5^\circ = 65^\circ$ , $\angle ACD = 170^\circ - 55^\circ = 115^\circ$ , $\angle BIA = \angle CID = 180^\circ - (65^\circ + 55^\circ) = 60^\circ$ , $\angle BIC = \angle AID = 180^\circ - 60^\circ = 120^\circ$ .

We can rewrite $\angle CAD$ and $\angle BDA$ in terms of $x$ . $\angle CAD = x - 55^\circ$ and $\angle BDA = 180^\circ - (120^\circ + x - 55^\circ) = 115^\circ - x$ .

Let us relabel $AB = BC = CD = a$ and $AD = b$ .

By Rule of Sines on $\triangle ACD$ and $\triangle ABD$ respectively, $\frac{\sin(\angle CAD)}{a} = \frac{\sin(\angle ACD)}{b}$ , and $\frac{\sin(\angle ABD)}{b} = \frac{\sin(\angle BDA)}{a}$

In a more convenient form, $\frac{\sin(x-55^\circ)}{a} = \frac{\sin(115^\circ)}{b} \implies \frac{a}{b} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$

and $\frac{\sin(65^\circ)}{b} = \frac{\sin(115^\circ-x)}{a} \implies \frac{a}{b} = \frac{\sin(115^\circ-x)}{\sin(65^\circ)}$

$\implies \frac{\sin(115^\circ-x)}{\sin(65^\circ)} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$

Now, by identity $\sin(\theta) = \sin(180^\circ-\theta)$ , $\sin(65^\circ) = \sin(115^\circ)$

Therefore, $\sin(115^\circ-x) = \sin(x-55^\circ).$ This equation is only satisfied by option $\boxed{\text {(C) } 85^\circ}$

Note: I'm pretty bad at Asymptote, if anyone could edit this and fill in the angles into the diagram, that would be pretty cool.

~Raghu9372

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

Contents

Problem

Solution 1

Solution 2

Solution 3(Using Trig.)

See also