Q1 of 32 Page 6

In Fig. 4.142, PA, QB and RC are each perpendicular to AC. Prove that .

(CBSE 2000)

We have, PA ⏊ AC, and RC ⏊ AC


Let AB = a and BC = b


In ΔCQB and ΔCPA


<QCB = <PCA (Common)


<QBC = <PAC (Each 90°)


Then, ΔCQB ~ ΔCPA (By AA similarity)


So, (Corresponding parts of similar triangle area proportion)


Or, -----------(i)


In ΔAQB and ΔARC


<QAB = <RAC (Common)


<ABQ = <ACR (Each 90°)


Then, ΔAQB ~ ΔARC (By AA similarity)


So, (Corresponding parts of similar triangle area proportion)


Or, -----------(ii)


Adding equation i & ii


=


Or, y () =


Or, y () = 1


Or, =

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