# PRACTICE PROBLEMS BASED ON BPT THEOREM

Practice Problems Based on bpt Theorem :

Here we are going to see some practice problems based on proportionality theorem

## Practice Problems Based on BPT Theorem

Question 1 :

In fig. if PQ || BC and PR || CD prove that (i)  AR/RD  =  AQ/AB

Solution :

PQ || BC

AP/PC  =  AQ/QB   ---(1)

PR || CD

AP/PC  =  AR/RD   ---(2)

(1)  =  (2)

AQ/QB  =  AR/RD

(ii)  QB/AQ  =  DR/AR

Solution :

Taking reciprocal on the first result, we get

QB/AQ  =  DR/AR

Question 2 :

Rhombus PQRB is inscribed in ΔABC such that ÐB is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus

Solution : In triangle ABC, BC is parallel to PQ.

Let side of rhombus be a cm.

Given AB = 12 cm. So AP = 12 – a

BC = 6 cm

We have AP / AB = PQ / BC

12 – a / 12 = a / 6

12 a = 72 – 6 a

18 a = 72

a = 72 / 18

a = 4 cm

PQ  =  4 cm and RB  =  4 cm

Question 3 :

In trapezium ABCD, AB || DC , E and F are points on non-parallel sides AD and BC respectively, such that EF || AB . Show that AE/ED  =  BF/FC

Solution : Let us join the vertices D and B which intersects the side EF at the point T. In triangle DAB,

DE/EA  =  DT/TB -----(1)

In triangle DBC

BT/TD  =   BF/FC -----(2)

Now let us take the reciprocal of (2) and equate them with (1). So we get

TD/BT  =  FC/BF

DE/DA  =  FC/BF

Take reciprocal on both sides, we get

DA/DE  =  BF/FC

Hence proved.

Question 4 :

In figure DE || BC and CD || EF. Prove that AD2 = AB×AF Solution :

In triangle ABC,

AF/FD  =  AE/EC -----(2)

(1)  =  (2)

Taking reciprocal on both side, so we get

Add 1 on both sides, we get

(DB/AD) + 1  =  (FD/AF ) + 1

Hence proved. After having gone through the stuff given above, we hope that the students would have understood, "Practice Problems Based on bpt Theorem".

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