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Let $ABCD$ be a cyclic quadrilateral with side lengths $AB=p,BC=q,CD=r,DA=s$.Show that $\dfrac{AC}{BD}=\dfrac{ps+qr}{pq+rs}$.

My work:
I have found out that this follows from Ptolemy's Second Theorem but cannot prove it.Please help! With or without Ptolemy is fine, I do not have any restriction.

Hawk
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3 Answers3

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As you've said this follows immidiatelly from the Second Ptolemy Therem. Here's the proof for it.

Note that in every triangle we have:

$$bc = 2Rh_{a}$$

This follows from the formula for area of triangle: $P = \frac{abc}{4R}$

Note: Vertex C and vertex D need to switch places.

So from $\triangle ABD$ and $\triangle BCD$ in the picture we have:

$$ad = 2Rh_1 \quad \quad \text { and } \quad \quad {bc = 2Rh_2}$$

Adding this we have:

$$ad + bc = 2Rh_1 + 2Rh_2$$

From the right trinagles, where $AK$ and $KD$ are hypothenyses we have:

$$h_1 = AK \cdot \sin w \quad \quad \text { and } \quad \quad {h_2 = KC \cdot \sin w}$$

Substitunting we have:

$$ad + bc = 2R \sin w (AK + KC) = 2R \cdot AC \sin w$$

Simularly we have:

$$ab + cd = 2R \cdot BD \sin (\pi - w)$$

But $\sin w = \sin (\pi - w)$, so we have:

$$\frac{ad + bc}{ab + cd} = \frac{2R \cdot AC \sin w}{2R \cdot BD \sin (\pi - w)} = \frac{AC}{BD}$$

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Stefan4024
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    @coffeemath As I've said under the diagram, points C and points D need to change places. – Stefan4024 Jan 26 '14 at 16:20
  • Stefan.. sorry about reading too quickly (comment deleted, and +1 for ans). – coffeemath Jan 26 '14 at 16:26
  • It was a very nice solution...but one thing I could not understand is that why $bc=2Rh_a$.Please explain that...It will be really helpful. – Hawk Jan 26 '14 at 18:48
  • @Hawk As I've mention use the formula for area of triangle: $P = \frac{abc}{4R}$, this is well-known formula and easy one to prove. Then just substitute $P = \frac{ah_a}{2}$ and we have:

    $$\frac{ah_a}{2} = \frac{abc}{4R} \implies 2Rh_a = bc$$

     – Stefan4024 Jan 26 '14 at 18:53
  • Oh..ok...I could not understand that you are talking about P=area...Thanks... – Hawk Jan 26 '14 at 18:55
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Recall that in a cyclic quadrilateral $\square ABCD$, opposite angles are supplementary (and, therefore, have a common sine). Also, as a consequence of the Law of Sines, a triangle $\triangle XYZ$ inscribed in a circle of diameter $d$ has side-lengths $$|YZ| = d \sin X \qquad |ZX| = d \sin Y \qquad |XY| = d \sin Z$$

For $\square ABCD$ inscribed in a circle of diameter $d$, then, we can write $$|BD| = d \sin A = d \sin C \qquad |AC| = d \sin B = d\sin D$$

Now,

$$\begin{align} ps+ qr &= \frac{2}{|BD|/d} \left(\; \frac{1}{2} p s \sin A + \frac{1}{2} qr \sin C \;\right) \\[4pt] &= \frac{2d}{|BD|} \left( |\triangle BAD| + |\triangle BCD| \right) \\[4pt] &= \frac{2d |\square ABCD|}{|BD|} \\[6pt] pq+rs &= \frac{2d|\square ABCD|}{|AC|} \end{align}$$ whence $$\frac{ps+qr}{pq+rs} = \frac{|AC|}{|BD|}$$

Blue
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  • I could not understand from 'For ABCD inscribed in a circle of diamter d.... please explain that a little more elaborately. – Hawk Jan 26 '14 at 18:50
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Area of a triangle of sides $ {a, b, c} $ inscribed in a circle of radius $ {R} $ is $ {{{ K}}= {\frac {abc} {4R}}} $

Now, let us consider a quadrilateral with sides ${ AB=p , BC=q , CD=r , DA=s }$.

enter image description here

In Figure 1 : Divide the quadrilateral into two triangles with area K1 and K2 respectively.Let the area of the quadrilateral be ${Q}$.

Using the formula for Area of Triangle,
$ {K1=\frac {BD.p.s}{4R}} $

${K2=\frac{BD.q.r}{4R}}$

So, ${ Q= K1+K2=\LARGE{{\frac {BD.(ps+qr)} {4R}}}}$ ....(1)

Similarly in Figure 2 :

${K3=\frac {AC.r.s} {4R}}$

${K4=\frac {AC.p.q} {4R}}$

So, ${ Q= K3+K4=\LARGE{{\frac {AC.(rs+pq)} {4R}}}}$ ....(2)

Now, from equation ${(1)}$ & ${(2)}$,

${{AC.(pq+rs)=BD.(ps+qr)}}$

Hence proved,

${\LARGE{{\frac {AC}{BD}} = {\frac {ps + qr}{pq + rs}}}}$

Abir Mukherjee
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