# A diagonal in a quadrilateral [closed]

A convex quadrilateral $ABCD$ with sides $AB=8$ cm, $BC=16$ cm, $CD=4$ cm and $AD=6$ cm is given. Find the diagonal $BD$ if the length is an integer.

## closed as off-topic by Namaste, Claude Leibovici, Magdiragdag, Especially Lime, HenrikApr 27 '17 at 8:38

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• Have you made an attempt at solving this? – Allan Apr 26 '17 at 19:45
• Okay. I'll do this. What will you be doing while I do this? – fleablood Apr 26 '17 at 19:46
• I use Triangle inequalities, but not sure about result. – vili Apr 26 '17 at 19:49
• pretty pictures don't compensate for lack of effort/context/attempts, etc. – Namaste Apr 27 '17 at 1:35
• @amWhy: the poster is telling that he worked around triangle inequalities. Does not seem to me that this post deserves to be put on hold more than many others. – G Cab Apr 27 '17 at 14:02

Use the triangle inequality on $\triangle ABD$ and $\triangle BCD$: a triangle exists with sides $a,b,c$ if and only if $a+b>c$, and cyclic permutations thereof. $6+8>BD$, $16>4+BD$. Wasn't that hard, was it?

• So is $ABCD$ is convex, $BD=13$ is the only possible integer value of $BD$. But why is it really attained by some configuration? Sorry for the nitpicking, but I guess it might be useful to state it clearly. (+1) by the way. – Jack D'Aurizio Apr 26 '17 at 19:59

Let's put $$\left\{ {\matrix{ {\mathop {AB}\limits^ \to = {\bf a}} & {\left| {\bf a} \right| = 8} \cr {\mathop {BC}\limits^ \to = {\bf b}} & {\left| {\bf b} \right| = 16} \cr {\mathop {CD}\limits^ \to = {\bf c}} & {\left| {\bf c} \right| = 4} \cr {\mathop {DA}\limits^ \to = {\bf d}} & {\left| {\bf d} \right| = 6} \cr } } \right.$$ by which $${\bf a + b + c + d} = {\bf 0}$$

Now we have that the diagonal equals $$diag = \mathop {BD}\limits^ \to = - {\bf a} - {\bf d = c} + {\bf b}$$ so that for its modulus we will have \eqalign{ & \left| {\mathop {BD}\limits^ \to } \right|^{\,2} = \left| {\bf a} \right|^{\,2} + \left| {\bf d} \right|^{\,2} + 2{\bf a} \cdot {\bf d} = \left| {\bf c} \right|^{\,2} + \left| {\bf b} \right|^{\,2} + 2{\bf c} \cdot {\bf b} = \cr & = 100 + 2{\bf a} \cdot {\bf d} = 272 + 2{\bf c} \cdot {\bf b} = \cr & = 100 + 2\left| {\bf a} \right|\left| {\bf d} \right|\cos \alpha = 272 + 2\left| {\bf c} \right|\left| {\bf b} \right|\cos \beta = \cr & = 100 + 96\cos \alpha = 272 + 128\cos \beta \cr}

Note that the angle $\alpha$ is the angle between the vectors $\mathop {AB}\limits^ \to$ and $\mathop {DA}\limits^ \to$, and therefore it is supplementary to the internal angle in $A$, and analogously for the angle $\beta$.

Since the cosine varies between $-1$ and $1$ we shall have \eqalign{ & \left\{ \matrix{ 100 - 96 \le \left| {\mathop {BD}\limits^ \to } \right|^{\,2} \le 100 + 96 \hfill \cr 272 - 128 \le \left| {\mathop {BD}\limits^ \to } \right|^{\,2} \le 272 + 128 \hfill \cr} \right.\quad \Rightarrow \quad \cr & \Rightarrow \quad 144 \le \left| {\mathop {BD}\limits^ \to } \right|^{\,2} \le 196\quad \Rightarrow \quad 12 \le \left| {\mathop {BD}\limits^ \to } \right| \le 14 \cr}

Therefore we get the following possible results

$$\matrix{ {\left| {\mathop {BD}\limits^ \to } \right| = 12} & {\cos \alpha = 44/96} & {\cos \beta = - 1} \cr {\left| {\mathop {BD}\limits^ \to } \right| = 13} & {\cos \alpha = 69/96} & {\cos \beta = - 103/128} \cr {\left| {\mathop {BD}\limits^ \to } \right| = 14} & {\cos \alpha = 1} & {\cos \beta = - 76/128} \cr }$$