  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: A
p2=b2+c2  B
1p2=1b2+1c2  C
pa=pb  D
p2=b2c2  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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