Class 10 Mathematics

# Triangle

## Exercise 6.6 (NCERT) Part 1

Question 1: In the given figure, PS is the bisector of ∠ QPR or Δ PQR. Prove that `(QS)/(SR)=(PQ)/(PR)`

**Solution:** Draw a line RT || SP; which meets QP extended up to QT.

∠ QPS = ∠ SPQ (given)

∠ SPQ = ∠ PRT (Alternate angles)

∠ QPS = ∠ PTR (Corresponding angles)

From these three equations, we have;

∠ PRT = ∠ PTR

Hence, in triangle PRT;

PT = PR (Sides opposite to equal angles) -----------(1)

Now; in triangles SQP and RQT;

∠ QPS = ∠ QTR (Corresponding angles)

∠ QSP = ∠ QRT (Corresponding angles)

Hence; Δ SQP ∼ ΔRQT (AAA criterion)

Hence, `(QS)/(SR)=(QP)/(PT)`

Or, `(QS)/(SR)=(QP)/(PR)`

Because PT = PR (from equation 1)

Question 2: In the given figure, D is a point on hypotenuse AC of Δ ABC, such that BD ⊥ AC, DM⊥ BC and DN ⊥ AN. Prove that:

(a) DM^{2} = DN.MC

**Solution:** DN || BC and DM || AB

DNMB is a rectangle because all the four angles are right angles.

Hence; DN = MB and DM = NB

In triangles DMB and CMD;

∠ DMB = ∠ CMD (Right angle)

∠ DBM = ∠ CDM

DM = DM

Hence; Δ DMB ∼ Δ CMD

Hence, `(DM)/(MB)=(CM)/(DM)`

Or, `DM^2=MB.MC`

Or, `DM^2=DN.MC`

(Because DN = MB)

(b) DN^{2} = DM.AN

**Solution:** In triangles DNB and AND;

∠ DNB = ∠ AND (Right angles)

∠ NDB = ∠ NAD

DN = DN

Hence; Δ DNB ∼ Δ AND

Hence, `(DN)/(NB)=(AN)/(DN)`

Or, `DN^2=NB.AN`

Or,`DN^2=DM.AN`

(Because DM = NB)

Question 3: In the given figure, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that AC^{2} = AB^{2} + BC^{2} + 2 BC.BD

**Solution:** In triangle ADB;

`AB^2 = AD^2 + BD^2` …….. (1)

In triangle ADC;

`AC^2 = AD^2 + DC^2`

Or, `AC^2 = AD^2 + (BD + BC)^2`

`= AD^2 + BD^2 + BC^2 + 2BD.BC` …….. (2)

Substituting the value of AB^{2} from equation (1) into equation (2), we get;

`AC^2 = AB^2 + BC^2 + 2BC.BD` proved

Question 4: In the given figure, ABC is triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC^{2} = AB^{2} + BC^{2} - 2BC.BD.

**Solution:** In triangle ABD;

`AB^2 = AD^2 + BD^2` …….. (1)

In triangle ADC;

`AC^2 = AD^2 + DC^2`

Or, `AC^2 = AD^2 + (BC – BD)^2`

`= AD^2 + BD^2 + BC^2 – 2BC.BD` ……… (2)

Substituting the value of AB^{2} from equation (1) in equation (2), we get;

`AC^2 = AB^2 + BC^2 – 2BC.BD` proved

Question 5: In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(a) `AC^2=AD^2+BC.DM+((BC)/(2))^2`

**Solution:** In triangle AMD;

`AD^2=AM^2+DM^2` ---------(1)

In triangle AMC:

`AC^2=AM^2+CM^2`

Or, `AC^2=AM^2+(DM+(BC)/(2))^2`

`=AM^2+DM^2+BC.DM+((BC)/(2))^2`---------(2)

Substituting the value of AD^{2} from equation (1) in equation (2), we get;

`AC^2=AD^2+BC.DM+((BC)/(2))^2`

Proved

(b) `AB^2=AD^2-BC.DM+((BC)/(2))^2`

**Solution:** In triangle ABM;

`AB^2=AM^2+BM^2`

`AB^2=AM^2+((BC)/(2)-DM)^2`

Or, `AB^2=AM^2+DM^2-BC.DM+((BC)/(2))^2`

Substituting the value of AD2 from equation (1) in equation (2), we get;

`AB^2=AD^2-BC.DM+((BC)/(2))^2`

(c) `AC^2+AB^2=2AD^2+1/2BC^2`

**Solution:** From question (a), we have;

`AC^2=AD^2+BC.DM+((BC)/(2))^2`

And from question (b), we have;

`AB^2=AD^2-BC.DM+((BC)/(2))^2`

Adding these two equations, we get;

`AC^2+AB^2=AD^2+BC.DM+((BC)/(2))^2``+ AD^2-BC.DM+((BC)/(2))^2`

Or, `AC^2+AB^2=2AD^2+1/2BC^2`

Proved

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Theorem (Part 1)

Theorem (Part 2)

Exercise 6.1

Exercise 6.2 (Part 1)

Exercise 6.2 (Part 2)

Exercise 6.3 (Part 1)

Exercise 6.3 (Part 2)

Exercise 6.3 (Part 3)

Exercise 6.4

Exercise 6.5 (Part 1)

Exercise 6.5 (Part 2)

Exercise 6.6 (Part 2)