Question

# In the given figure, $MGN$ is a triangle and $MP$ is perpendicular to $GN$. If $MG = a$ units, $MN = b$ units, $GP = c$ units and $PN = d$ units, prove that $\dfrac{{(a - b)}}{{(c - d)}} = \dfrac{{(c + d)}}{{(a + b)}}.$

In the given figure, $\Delta GMP$ is a right angled triangle. So using Pythagoras theorem, we’ll get:
${(MP)^2} + {c^2} = {a^2}, \\ \Rightarrow {(MP)^2} = {a^2} - {c^2} .....(i) \\$
$\Delta MNP$ is also a right angled triangle. So using Pythagoras theorem again, we’ll get:
${(MP)^2} + {d^2} = {b^2}, \\ \Rightarrow {(MP)^2} = {b^2} - {d^2} .....(ii) \\$
From equation $(i)$ and $(ii)$, we have:
$\Rightarrow {a^2} - {c^2} = {b^2} - {d^2}, \\ \Rightarrow {a^2} - {b^2} = {c^2} - {d^2}, \\ \Rightarrow (a + b)(a - b) = (c + d)(c - d), \\ \Rightarrow \dfrac{{(a - b)}}{{(c - d)}} = \dfrac{{(c + d)}}{{(a + b)}}. \\$