Question

In the given figure, PA, QB and RC are each perpendicular to AC. Prove that 1/x + 1/z = 1/y.

Answer 1

SOLUTION :  

Given :  PA ⊥ AC, QB ⊥ AC and RC ⊥ AC

Let, AB = a and BC = b

In ∆CQB and ∆CPA

∠QCB = ∠PCA [Common]

∠QBC = ∠PAC [Each 90°]

∆CQB ~ ∆CPA [By AA similarity]

∴QB/PA = CB/CA …………..(1)

[Corresponding parts of similar triangles are proportional]

y/x= b/(a+b)

In ∆AQB and ∆ARC

∠QAB = ∠RAC  [Common]

∠ABQ = ∠ACR   [Each 90°]

∆AQB ~ ∆ARC [By AA similarity]

∴QB/RC = AB/AC  

[Corresponding parts of similar triangles are proportional]

y/z = a/(a+b)…………..(2)

On Adding eq. 1 & 2

y/x + y/z = b/(a+b) + a/(a+b)

y(1/x + 1/z) = (b+a)/(a+b)

y(1/x + 1/z) = 1

1/x + 1/z = 1/y

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Answer 2

ANSWER:--

 Given :  PA ⊥ AC, QB ⊥ AC and RC ⊥ AC Let, AB

= a and BC

= b In ∆CQB and ∆CPA ∠QCB

= ∠PCA [Common] ∠QBC = ∠PAC [Each 90°] ∆CQB ~ ∆CPA [By AA similarity] ∴QB/PA

= CB/CA …………..

(1) [Corresponding parts of similar triangles are proportional] y/x= b/

(a+b) In ∆AQB and ∆ARC ∠QAB

= ∠RCA   [Common] ∠ABQ

= ∠ACR   [Each 90°] ∆AQB ~ ∆ARC [By AA similarity]

∴QB/RC = AB/AC  [Corresponding parts of similar triangles are proportional] y/z = a/

(a+b)…………..(2) On Adding eq.

1 & 2 y/x + y/z

= b/(a+b) + a/(a+b) y(1/x + 1/z)

= (b+a)/(a+b) y(1/x + 1/z)

= 1 1/x +1/z = 1/y

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