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Question

In a right angled $$\triangle ABC$$, right angled at $$A$$, if $$AD\bot BC$$ such that $$AD=p$$, if $$BC=a, CA=b$$ and $$AB=c$$, then:
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A
p2=b2+c2
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B
1p2=1b2+1c2
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C
pa=pb
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D
p2=b2c2
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Solution

The correct option is B $$\cfrac { 1 }{ { p }^{ 2 } } =\cfrac { 1 }{ { b }^{ 2 } } +\cfrac { 1 }{ { c }^{ 2 } } $$
Given, In $$\triangle ABC $$, $$\angle A = 90 $$ and $$ AD \perp BC $$
In $$\triangle ABC$$, 

$$\angle BAC + \angle ABC + \angle ACB = 180 $$

$$90 + \angle ABC + \angle ACB = 180 $$

$$\angle ABC + \angle ACB = 90 $$ (I)
In $$\triangle CAD$$,

$$\angle CAD + \angle ACD + \angle ADC = 180 $$

$$\angle CAD + \angle ACD + 90 = 180 $$

$$\angle CAD + \angle ACD = 90 $$..(II)

Equating (I) and (II),
$$\angle ABC + \angle ACB = \angle CAD + \angle ACD $$
$$\angle ABC = \angle CAD $$...(III)
Similarly, $$\angle ACB = \angle BAD $$...(IV)
Now, In $$\triangle$$s, ABC and DAC

$$\angle ABC = \angle CAD $$ ..(From III)

$$\angle BAC = \angle CDA$$ (Each $$90^{\circ})$$

$$\angle ACD = \angle ACB $$ (Common angle)

Thus, $$\triangle ABC \sim \triangle DAC$$ (AAA rule)

Thus, $$\dfrac{AC}{DC} = \dfrac{BC}{AC}$$ (Sides of similar triangles are in proportion)

$$AC^2 = BC \times DC $$... (V)
Similarly, $$\triangle ABC \sim \triangle DBA$$

and $$\dfrac{AB}{BD} = \dfrac{BC}{AB}$$ (Sides of similar triangles are in proportion)

$$AB^2 = BD \times BC $$  ....(VI)
Similarly, $$\triangle ABD \sim \triangle CAD$$ (AAA rule)

Thus, $$\dfrac{AD}{CD} = \dfrac{BD}{AD}$$ (Sides of similar triangles are in proportion)

$$AD^2 = BD \times CD $$
Now, 
$$\dfrac{1}{AB^2} + \dfrac{1}{AC^2} = \dfrac{1}{BC \times DC} + \dfrac{1}{BD \times BC}$$

$$\dfrac{1}{AB^2} + \dfrac{1}{AC^2} = \dfrac{1}{BC} (\dfrac{1}{DC} + \dfrac{1}{BD})$$

$$\dfrac{1}{AB^2} + \dfrac{1}{AC^2} = \dfrac{1}{BC} (\dfrac{BD + DC}{BD \times DC})$$

$$\dfrac{1}{AB^2} + \dfrac{1}{AC^2} = \dfrac{1}{BC} (\dfrac{BC}{BD \times DC})$$

$$\dfrac{1}{AB^2} + \dfrac{1}{AC^2} = \dfrac{1}{BD \times DC}$$

$$\dfrac{1}{AB^2} + \dfrac{1}{AC^2} = \dfrac{1}{AD^2}$$

$$\dfrac{1}{c^2} + \dfrac{1}{b^2} = \dfrac{1}{p^2}$$

Mathematics

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