
5. A trapezium $A B C D$, in which $A B$ is parallel to $C D$, is inscribed in a circle with centre $O$. Suppose the diagonals $A C$ and $B D$ of the trapezium intersect at $M$, and $O M=2$.

(a) If $\angle A M B$ is $60^{\circ}$, determine, with proof, the difference between the lengths of the parallel sides.

(b) If $\angle A M D$ is $60^{\circ}$, find the difference between the lengths of the parallel sides.


## Solution:

Suppose $\angle A M B=60^{\circ}$. Then $A M B$ and $C M D$ are equilateral triangles. Draw $O K$ perpendicular to $B D$.(see Fig.1) Note that $O M$ bisects $\angle A M B$ so that $\angle O M K=$
$30^{\circ}$. Hence $O K=O M / 2=1$. It follows that $K M=\sqrt{O M^{2}-O K^{2}}=\sqrt{3}$. We also observe that

$$
A B-C D=B M-M D=B K+K M-(D K-K M)=2 K M
$$

since $K$ is the mid-point of $B D$. Hence $A B-C D=2 \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_627ae6532caab02b227eg-4.jpg?height=542&width=523&top_left_y=409&top_left_x=411)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_06_05_627ae6532caab02b227eg-4.jpg?height=507&width=572&top_left_y=446&top_left_x=1053)

Fig. 2

Suppose $\angle A M D=60^{\circ}$ so that $\angle A M B=120^{\circ}$. Draw $P Q$ through $O$ parallel to $A C$ (with $Q$ on $A B$ and $P$ on $B D$ ). (see Fig.2) Again $O M$ bisects $\angle A M B$ so that $\angle O P M=\angle O M P=60^{\circ}$. Thus $O M P$ is an equilateral triangle. Hence diameter perpendicular to $B D$ also bisects $M P$. This gives $D M=P B$. In the triangles $D M C$ and $B P Q$, we have $B P=D M, \angle D M C=120^{\circ}=\angle B P Q$, and $\angle D C M=$ $\angle P B Q$ (property of cyclic quadrilateral). Hence $D M C$ and $B P Q$ are congruent so that $D C=B Q$. Thus $A B-D C=A Q$. Note that $A Q=K P$ since $K A Q P$ is a parallelogram. But $K P$ is twice the altitude of triangle $O P M$. Since $O M=2$, the altitude of $O P M$ is $2 \times \sqrt{3} / 2=\sqrt{3}$. This gives $A Q=2 \sqrt{3}$.

## Alternate Solution:

Using some trigonometry, we can get solutions for both the parts simultaneously. Let $K, L$ be the mid-points of $A B$ and $C D$ respectively. Then $L, M, O, K$ are collinear (see Fig. 3 and Fig.4). Let $\angle A M K=\theta(=\angle D M L)$, and $O M=d$. Since $A M B$ and $C M D$ are similar triangles, if $M D=M C=x$ then $M A=M B=k x$ for some positive constant $k$.

Now $M K=k x \cos \theta, M L=x \cos \theta$, so that $O K=|k x \cos \theta-d|$ and $O L=x \cos \theta+d$. Also $A K=k x \sin \theta$ and $D L=x \sin \theta$. Using

$$
A K^{2}+O K^{2}=A O^{2}=D O^{2}=D L^{2}+O L^{2}
$$

we get

$$
k^{2} x^{2} \sin ^{2} \theta+(k x \cos \theta-d)^{2}=x^{2} \sin ^{2} \theta+(x \cos \theta+d)^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_627ae6532caab02b227eg-5.jpg?height=563&width=571&top_left_y=106&top_left_x=384)

Fig. 3

![](https://cdn.mathpix.com/cropped/2024_06_05_627ae6532caab02b227eg-5.jpg?height=523&width=570&top_left_y=145&top_left_x=1054)

Fig. 4

Simplification gives

$$
\left(k^{2}-1\right) x^{2}=2 x d(k+1) \cos \theta
$$

Since $k+1>0$, we get $(k-1) x=2 d \cos \theta$. Thus

$$
\begin{aligned}
A B-C D=2(A K-L D) & =2(k x \sin \theta-x \sin \theta) \\
& =2(k-1) x \sin \theta \\
& =4 d \cos \theta \sin \theta \\
& =2 d \sin 2 \theta
\end{aligned}
$$

If $\angle A M B=60^{\circ}$, then $2 \theta=60^{\circ}$. If $\angle A M D=60^{\circ}$, then $2 \theta=120^{\circ}$. In either case $\sin 2 \theta=\sqrt{3} / 2$. If $d=2$, then $A B-C D=2 \sqrt{3}$, in both the cases.
