Question

In the given figure, each one of PA, QB and RC is perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, show that `1/x + 1/y = 1/z`.

Answer 1

Since PA, QB and RC are all perpendicular to AC, we have QB || PA and QB || RC.

1. In triangle PAC, QB || PA, so triangles QBC and PAC are similar.

Hence `z/x = (BC)/(AC) = b/(a + b)`.   ...(1)

2. In triangle RCA, QB || RC, so triangles QBA and RCA are similar.

Hence `z/y = (AB)/(AC) = a/(a + b)`.   ...(2)

Add (1) and (2):

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

= `(a + b)/(a + b) = 1`

Factor z:

`z(1/x + 1/y) = 1` 

⇒ `1/x + 1/y = 1/z`

Thus `1/x + 1/y = 1/z`.