2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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:

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

$\endgroup$
  • 1
    $\begingroup$ @coffeemath As I've said under the diagram, points C and points D need to change places. $\endgroup$ – Stefan4024 Jan 26 '14 at 16:20
  • $\begingroup$ Stefan.. sorry about reading too quickly (comment deleted, and +1 for ans). $\endgroup$ – coffeemath Jan 26 '14 at 16:26
  • $\begingroup$ 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. $\endgroup$ – Hawk Jan 26 '14 at 18:48
  • $\begingroup$ @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: $$\frac{ah_a}{2} = \frac{abc}{4R} \implies 2Rh_a = bc$$ $\endgroup$ – Stefan4024 Jan 26 '14 at 18:53
  • $\begingroup$ Oh..ok...I could not understand that you are talking about P=area...Thanks... $\endgroup$ – Hawk Jan 26 '14 at 18:55
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I could not understand from 'For ABCD inscribed in a circle of diamter d.... please explain that a little more elaborately. $\endgroup$ – Hawk Jan 26 '14 at 18:50
0
$\begingroup$

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,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.