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Hint: Here the given triangle is a right angled triangle, we can use Pythagoras theorem.

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)}}. \\

$

This is the required proof.

Note: In any triangle, perpendicular drawn from a vertex to the opposite side always divides the triangle into two right angled triangles. We can use Pythagoras theorem for both the triangles separately.

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)}}. \\

$

This is the required proof.

Note: In any triangle, perpendicular drawn from a vertex to the opposite side always divides the triangle into two right angled triangles. We can use Pythagoras theorem for both the triangles separately.

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