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)`
Answer: 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
Answer: 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
Answer: 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 AC2 = AB2 + BC2 + 2 BC.BD
Answer: 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 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.
Answer: 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 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`
Answer: 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`
Answer: 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`
Answer: 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