# 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) DM2 = 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) DN2 = DM.AN

Solution: In triangles DNB and AND;

∠ DNB = ∠ AND (Right angles)

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 AC2 = AB2 + BC2 + 2 BC.BD AB^2 = AD^2 + BD^2 …….. (1)

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 AB2 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 AC2 = AB2 + BC2 - 2BC.BD. Solution: In triangle ABD;

AB^2 = AD^2 + BD^2 …….. (1)

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 AB2 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 AD2 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