In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on line CD such that DQ = 4 and QC = 1. Find sin angle PBQ.

Answer 1

Let $X$ be the point on line $AD$ such that $PX \perp AD$ and $Y$ be the point on line $AD$ such that $QY\perp AD$, as shown below.

[asy]
pair A,B,C,D,P,Q,X,Y;
A=(0,0);
B=(0,1);
C=(1,1);
D=(1,0);
P=(0,4/5);
Q=(1,4/5);
X=(0,4/5);
Y=(1,4/5);
draw(A--B--C--D--cycle);
draw(P--X);
draw(Q--Y);
draw(P--Q);
label("$A$",A,SW);
label("$B$",B,W);
label("$C$",C,NE);
label("$D$",D,SE);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$X$",X,E);
label("$Y$",Y,E);
[/asy]

Since $BP=4$ and $PC=1$, we have $BC=BP+PC=4+1=5$. Since $DQ=4$ and $QC=1$, we have $DC=DQ+QC=4+1=5$. Therefore, $BC=CD=5$, so $\triangle BCD$ is an isosceles right triangle.

Since $\triangle BCD$ is an isosceles right triangle, we have $\angle BCD = 90^\circ$. Because $\angle PBX$ and $\angle QBY$ are each complementary to $\angle BCD$, we have $\angle PBX = \angle QBY$ and $\angle XBP = \angle YBQ$. Therefore, $\triangle PBX\sim \triangle QBY$.

Since $\triangle PBX\sim \triangle QBY$, we have $\frac{BX}{BY}=\frac{BP}{BQ}=\frac{PB}{QB}$. Also, $\triangle BXP\sim \triangle BQY$, so $\frac{BX}{BQ}=\frac{BP}{BY}$. Therefore, \[\frac{PB}{QB}=\frac{BX}{BY}=\frac{BP}{BQ}=\frac{PB}{BY}.\] Since $BY\neq 0$, we have $QB=BQ$. Therefore, $\triangle BQP$ is an isosceles triangle, so $\angle PBQ = \angle BQP$.

After constructing the altitude from $P$ to $\overline{BQ}$, we have the following diagram:

[asy]
pair A,B,C,D,P,Q,X,Y;
A=(0,0);
B=(0,1);
C=(1,1);
D=(1,0);
P=(0,4/5);
Q=(1,4/5);
X=(0,4/5);
Y=(1,4/5);
draw(A--B--C--D--cycle);
draw(P--X);
draw(Q--Y);
draw(P--Q);
draw(B--Q--P,dashed);
label("$A$",A,SW);
label("$B$",B,W);
label("$C$",C,NE);
label("$D$",D,SE);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$X$",X,E);
label("$Y$",Y,E);
label("$4$",(P+X)/2,W);
label("$1$",(Q+Y)/2,E);
[/asy]

Since $\triangle PBY \sim \triangle QBX$, we have $\frac{BY}{YQ}=\frac{BP}{PX}$. Since $BQ=BP$ (as shown above), we have $\triangle PBQ\cong \triangle PBX$. Therefore, $\angle PBQ = \angle PBX$, so $\angle PBQ = \angle BXP$.

Finally, since $\triangle PBX\sim \triangle BCD$, we have $\frac{XP}{BC}=\frac{BX}{BD}$. Since $BC=5$ and $BD=5$, this simplifies to $\frac{XP}{5}=\frac{BX}{5}$. Because $XP=BP=4$, we have $BP=BX$, so $\angle PBQ = \angle BXP = \angle PBX=45^\circ$. Therefore, $\sin \angle PBQ = \sin \left(45^\circ\right)=\boxed{\frac{\sqrt2}{2}}$.