Q. 204.4( 9 Votes )

In the given figu

Answer :

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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