# Triangle

## Exercise 6.2 Part 2

Question 5: In the given figure, DE || OQ and DF || OR. Show that EF || QR. Solution: In Δ PQO and Δ PED;

(PE)/(EQ)=(ED)/(QO)=(PD)/(DO)

(Because these are similar triangles, as per Basic Proportionality theorem.)

Similarly, in Δ PRO and Δ PFD;

(PF)/(FR)=(FD)/(RO)=(PD)/(DO)

From above two equations, it is clear;

(PE)/(EQ)=(PF)/(FR)

Hence;

(PE)/(EQ)=(EF)/(QR)

Hence, EF || QR proved.

Question 6: In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. Solution: In Δ OPQ and Δ OAB;

(OA)/(AP)=(OB)/(BQ)

(Because these are similar triangles as per BPT.)

Similarly, in Δ OPR and Δ OAC;

(OA)/(OP)=(OC)/(CR)

From above two equations, it is clear;

(OB)/(BQ)=(OC)/(CR)

Hence; BC || QR proved

Question 7: Using theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. Solution: PQR is a triangle in which DE || QR. Line DE intersects PQ at D so that PD = DQ

To Prove: PE = ER

In Δ PQR and Δ PDE;

(PD)/(DQ)=(PE)/(ER)

(Because as per BPT, these are similar triangles.)

(PD)/(DQ)=1 given

Hence, (PE)/(ER)=1

Hence; E is the midpoint of PR proved.

Question 8: Using theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Solution: The figure from the previous question can be used to solve this question.

ABC is a triangle in which D and E are mid points of PQ and PR respectively.

To Prove: DE || QR

In Δ PQR and Δ PDE;

(PD)/(DQ)=(PE)/(ER)=1 given Hence, as per BPT these triangles are similar.

Hence; DE || QR proved.

Question 9: ABCD is a trapezium in which AB || CD and its diagonals intersect each other at the point O. show that; (AO)/(BO)=(CO)/(DO) Solution: Draw a line EF || CD which is passing through O.

In Δ ABC and Δ EOC;

These are similar triangles as per BPT.

(AE)/(EC)=(BO)/(OC)

Similarly, in Δ BOD and Δ FOD;

(BF)/(FD)=(AO)/(OD)

In Δ ABC and Δ BAD;

(BO)/(OC)=(AO)/(OD)

(Because diagonals of a trapezium divide each other in same ratio)

From above three equations, it is clear;

(AE)/(EC)=(BF)/(FD)

Hence, Δ ABC ~ Δ BAD

Using the third equation;

(BO)/(OC)=(AO)/(OD)

Or, (AO)/(BO)=(CO)/(DO)

Question 10: The diagonals of a quadrilateral ABCD intersect each other at point O such that (AO)/(BO)=(CO)/(DO) Show that ABCD is a trapezium.

Solution: This question can be proved by using the figure in previous question. Since diagonals are dividing each other in the same ratio hence, ABCD is a trapezium; proved.