Chapter 12.6, Problem 17WE

To determine

To prove $d_1^2+d_2^2=2\left(a^2+b^2\right)$

Explanation of Solution

Given information :

  

We shall use law of cosine in triangle ADB.

Where DB is a diagonal $d_1$

Cosine formula states that,

  $\cos A=\frac{b^2+c^2-a^2}{2bc}$

Putting values from figure we get,

  $\begin{array}{l}\cos \theta =\frac{a^2+b^2-{\left(d_1\right)}^2}{2\left(a\right)\left(b\right)}\\ 2ab\cos \theta =a^2+b^2-{\left(d_1\right)}^2\\ d_1^2=a^2+b^2-2ab\cos \theta \end{array}$

Now let AC be $d_2$

Similarly,

  $\begin{array}{l}\cos \left({180}^{\circ }-\theta \right)=\frac{a^2+b^2-{\left(d_2\right)}^2}{2\left(a\right)\left(b\right)}\\ -2ab\cos \theta =a^2+b^2-{\left(d_2\right)}^2\\ d_2^2=a^2+b^2+2ab\cos \theta \end{array}$

Now we add both the equations we get,

  $\begin{array}{l}{d}_1^2+d_2^2=a^2+b^2\cancel{-2ab\cos \theta }+a^2+b^2+\cancel{2ab\cos \theta }\\ d_1^2+d_2^2=2\left(a^2+b^2\right)\end{array}$

Hence proved.