Given: PA ⊥ AC, QB ⊥ AC and RC ⊥ AC

AP = x, QB = z, RC = y, AB = a and BC = b

To show:

In ∆ PAC, we have

QB ∥ PA

So, by Basic Proportionality theorem, we have

……….(i)

In ∆ ARC, we have

QB ∥ RC

So, by Basic Proportionality theorem, we have

……….(ii)

Now, Consider ∆ PAC and ∆ QBC,

∠PCA = ∠QCB [Common angle]

[By (i)]

So, by SAS criterion,

∆ PAC ∼ ∆ QBC

⇒ Ratio of all the corresponding sides of ∆ ABC and ∆ DEF are equal.

……….(iii)

Now, consider ∆ ARC and ∆ AQB,

∠RAC = ∠QAB [Common angle]

[By (ii)]

So, by SAS criterion,

∆ ARC ∼ ∆ AQB

⇒ Ratio of all the corresponding sides of ∆ ARC and ∆ AQB are equal.

……….(iv)

Now, adding (iii) and (iv), we get