Question

[2 ] {149} PA,QB AND RC are all perpendicULAR to AC. prove that 1/x+1/z =1/y

Answer 1

Solution:-
In Δ PAC and Δ QBC,
∠ PCA = ∠ QCB       (Common angle)
∠ PAC = ∠ QBC       (Right angle each)
⇒ Δ PAC ~ Δ QBC    (AA Similarity)
⇒ PA/QB = AC/BC    (Similar triangles have proportional sides)
⇒ x/y = AC/BC ⇒ y/x = BC/AC .....(1)
In Δ RCA and Δ QBA,
∠ RAC = ∠ QAB        (Common angle)
∠ RCA = ∠ QBA        (Right angle each)
⇒ Δ RCA ~ Δ QBA     (AA Similarity)
⇒ RC/QB = AC/AB 
⇒ z/y = AC/AB ⇒ y/z = AB/AC ......(2)
Adding (1) and (2), we get
y/x + y/z = (BC + AC)/AC = AC/AC = 1
⇒ y/x + y/z = 1
On multiplying both sides by 1/y, we get
1/x + 1/z = 1/y
Hence proved.

Answer 2

$\huge{\pink{\underline{\ulcorner{\star\: Solution \: \star}\urcorner}}}$

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$\red{\underline{Given\: \colon}}$

PA , QB , RC are perpendicular to AC

$\red{\underline{To \: Prove\: \colon}}$}

$\frac{1}{x}+\frac{1}{z} = \frac{1}{y}$

$\red{\underline{Solution\:\colon}}$}

$In\:\green{\triangle{CBQ}} \: \& \: \triangle{CAP}$

$\angle{1} = \angle{1}........(common) \\\\ \angle{6} = \angle{3}$

$\green{\triangle{CBQ}}\: \sim\: \triangle{CAP}$

AA - similarity rule

$\frac{CB}{CA}= \frac{BQ}{AP}$.....(1)

$\large{\frac{b}{b+a} = \frac{y}{x}}$.........[i]

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$In \:\red{\triangle{ABQ}} \: \& \: \triangle{ARC}$

$\angle{2} = \angle{2}........(common) \\\\ \angle{5} = \angle{4}.........(each\: 90\degree)$

$\red{\triangle{ABQ}}\: \sim\: \triangle{ACR}$

AA - similarity rule

$\frac{AB}{AC}= \frac{BQ}{CR}$.....(2)

$\large{\frac{a}{a+b} = \frac{y}{z}}$.........[ii]

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Now,

Add equation [i] & [ii]

$\frac{b}{b+a}\: + \: \frac{a}{a+b}= \frac{y}{x}\: + \: \frac{y}{z}\\\\ \frac{a+b}{b+a} = (y)(\frac{1}{x} + \frac{1}{z}) \\\\ \frac{\cancel{a+b}}{\cancel{a+b}} = (y)(\frac{1}{x} + \frac{1}{z}) \\\\ 1 = y(\frac{1}{x} + \frac{1}{z}) \\\\ \frac{1}{y}= \frac{1}{x} + \frac{1}{z}$

$\huge{\red{\boxed{\frac{1}{x}+\frac{1}{z} = \frac{1}{y}}}}$

Hence Proved