# Triangle

## NCERT Exercise 6.6

### 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