Question

ABC is a right angled triangle , \[\left| \!{\underline {\,
  C \,}} \right. = {90^ \circ }\]. If P is the perpendicular from C to AB and a,b,c have the usual meaning , then prove that \[\dfrac{1}{{{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}\].

Answer 1

Hint:As the given triangle is a right-angled one, we can apply Pythagoras theorem to the sides of the triangle. As the line is drawn as a perpendicular line to the hypotenuse, there will be two inner right-angled triangles formed. We can find the area of those two inner triangles and sum it and equate it to the whole triangle area.

Complete step by step answer:
$\Rightarrow$ Given ABC is a right-angled triangle with the right angle at C.
$\Rightarrow$ Therefore \[\left| \!{\underline {\,C \,}} \right. = {90^ \circ }\].
Let P is the perpendicular from C to AB and it intersects AB at D . given the length of the side AB = c, BC = a, CA = b.
$\Rightarrow$ Therefore CD = p .
$\Rightarrow$ We know that the Area of the triangle is equal to \[\dfrac{1}{2}\]( base of a triangle ) ( height of a triangle ) sq units.
$\Rightarrow$ By taking BC as base
Area of triangle ABC = \[\dfrac{1}{2}\]x BC x AC sq .units .
 = \[\dfrac{1}{2}\]x a x b sq . units .
  = \[\dfrac{1}{2}\]ab sq units .
Similarly
$\Rightarrow$ By taking AB as a base we get
Area of triangle ABC = \[\dfrac{1}{2}\]x AB x CD sq units .
= \[\dfrac{1}{2}\]cp sq units .
Therefore ab = pc .
$\Rightarrow$ That shows that \[\dfrac{{ab}}{p}\] = c ………….. ( 1 ) .
We know that in a right-angled triangle we can apply Pythagoras theorem.
Pythagoras theorem states that some of the squares on the lengths of a right angle is equal to the square on the hypotenuse.
$\Rightarrow$ It is written as \[{a^2}\]+ \[{b^2}\]= \[{c^2}\].
$\Rightarrow$ By applying Pythagoras theorem for the above triangle we get
  \[{a^2}\]+ \[{b^2}\]= \[{c^2}\]
  By replacing c with \[\dfrac{{ab}}{p}\] we get
\[{a^2}\]+ \[{b^2}\]= \[{\left( {\dfrac{{ab}}{p}} \right)^2}\]
$\Rightarrow$ Therefore by solving we get \[\dfrac{1}{{{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}\] .
Hence proved.

Note:
We can also prove the above question in another method. As we know that the whole hypotenuse of the triangle ABC is the sum of two sides from two inner triangles.
So we can find those two sides in terms of p, a and b and equate it to the square root of \[{a^2}\]+ \[{b^2}\] which is C. Then we will get an equation in terms of a, b and p.