# 10th CBSE math solution for exercise 6.2 part 2

This page 10th CBSE math solution for exercise 6.2 part 2 is going to provide you solution for every problems that you find in the exercise no 6.2

## 10th CBSE math solution for exercise 6.2

(5) In fig 6.20 DE **∥ OQ and DF ∥ OR .show that EF ∥ QR**

In triangle PQO

DE **∥ OQ**

(PE/EQ) = (PD/DO)----(1)

In triangle POR

DE **∥ OQ**

(PF/FR) = (PD/DO) ------(2)

(PE/EQ) = (PF/FR) by using inverse of BPT theorem

EF is parallel to QR

**(6) In fig 6.21, 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 triangle OPQ**

**AB is parallel to PQ**

**(OA/AP) = (OB/BQ) ---(1)**

** In triangle OPR**

** AC is parallel to PR**

** (OA/AP) = (OC/CR)**** ---(2)**

**(1) = (2)**

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

**by using converse of BPT theorem **

**BC is parallel to QR**

**(7)
Using theorem 6.1,prove that a line drawn through the midpoint of one
side of a triangle parallel to another side bisects the third side.**

**Solution:**

**Draw OM parallel to AB meeting BC at M. **

**In triangle ACB **

**OM is parallel to AB**

**(OC/OA) = (CM/MB) ----(1)**

** similarly in triangle BDC**

**OM is parallel to CD**

**(BM/MC) = (OB/OD)**

**taking reciprocal on both sides **

** (CM/MB) = (OD/OB) ----(2)**

** ****(OC/OA) =****(OD/OB)**

** (OC/OD) = (OA/OB)**

** hence proved.**

**(8) Using theorem 6.2 prove that the line joining the midpoint of any two sides of a triangle is parallel to the third side**

∴ AD=DB

⇒ AD/BD = 1 ---- (i)

Also, E is the mid-point of AC (Given)

∴ AE=EC

⇒AE/EC = 1 [From equation (i)]

From equation (i) and (ii), we get

AD/BD = AE/EC

Hence, DE || BC [By converse of Basic Proportionality Theorem]

**(9) ABCD is a trapezium in which AB ****∥ DC and its diagonals intersect each other at the point O. Show that (AO/BO) = (CO/DO)**

In ΔADC, we have

OE || DC (By Construction)

∴ AE/ED = AO/CO ...**(i)** [By using Basic Proportionality Theorem]

In ΔABD, we have

OE || AB (By Construction)

∴ DE/EA = DO/BO ...**(ii)** [By using Basic Proportionality Theorem]

From equation **(i)** and **(ii)**, we get

AO/CO = BO/DO

⇒ AO/BO = CO/DO

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