Question

triangle ABC is a right angled at C. If p is the length of the perpendicular from C to AB and a, b,c are the lengths

of the sides opposite angleA, angleB and angleC respectively, then prove that 1/p square=1/a square +1/b square​

Answer 1

Answer:

In ∆ADC and ∆ACB,

Angle A = Angle A (Common)

Angle ADC = Angle ACB (90° each)

.•. ∆ADC ~ ∆ACB (AA Similarity)

AC/AB = DC/BC (By CPST)

$\frac{b}{c} = \frac{p}{a}$

(From the diagram)

Do the reciprocal,

$\frac{c}{b} = \frac{a}{p}$

Bring a on the L.H.S,

$\frac{c}{ab} = \frac{1}{p}$

On Squaring Both Sides,

$\frac{ {c}^{2} }{ {a}^{2} {b}^{2} } \: = \frac{1}{ {p}^{2} }$

$\frac{ {a}^{2} + {b}^{2} }{ {a}^{2} {b}^{2} } = \frac{1}{ {p}^{2} } \:$

[Because, By Pythagoras Theorem,

${c}^{2} = {a}^{2} + {b}^{2}$

]

By Splitting the denominator to each numerator,

$\frac{ {a}^{2} }{ {a}^{2} {b}^{2} } + \frac{ {b}^{2} }{ {a}^{2} {b}^{2} } = \frac{1}{ {p}^{2} }$

Now by cancelling the same terms in the numerator and denominator we get our answer :

$\frac{1}{ {p}^{2} } = \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} }$

Hence Proved.

Hope this helps.

Have a nice day =)