# In the given figu

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

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