# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Claude Leibovici, Magdiragdag, Especially Lime, Henrik
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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

## 2 Answers

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

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