10类NCERT解决方案-第6章三角形-练习6.6

问题1.在图中，PS是ΔPQR的∠QPR的二等分线。证明$\frac{QS}{SR} = \frac{PQ}{PR}.$

解决方案：

Given: PS is angle bisector of ∠QPR.

To prove: $\frac{QS}{SR} = \frac{PQ}{PR}$

Construction: Draw RT∥ST to cut OP provided at T.

Proof: since PS∥TR and PR cuts them, hence we have,

∠1 = ∠3           ——[alternate interior angle]  1

∠1 = ∠4           ——[corresponding angle]        2

But ∠1 = ∠2     [given]

From 1 and 2

∴∠3=∠4  

In ∆PTR,

PT = PR    [sides opposite to equal angle are equal]  3

Now In ∆QRT, we have

$\frac{QS}{SR} = \frac{PQ}{PT}$                    ———-[by B.P.T]

$\frac{QS}{SR} = \frac{PQ}{PR}$   ———[from 3]

问题2。在图中，D是ΔABC的斜边AC上的一点，如BD⊥AC，DM⊥BC和

DN⊥AB。证明：

(i)DM ² = DN.MC

(ii)DN ² = DM.AN

解决方案：

Given: BD ⊥ AC, DM ⊥ BC and DN ⊥ AB

To prove:  

(i) DM² = DN.MC

(ii) DN² = DM.AN

Proof: we have AB⊥BC and DM⊥BC

So, AB || DM

Similarly, we have CB⊥AB and DN ⊥ AB

Therefore, CB || DN

Hence, BMD DN is a triangle

∴BM=DN or DM=ND  

In ∆BMD,

∠1+ ∠2+ ∠DMB=180           ———–[angle sum property of triangle]

∠1+ ∠2+90=180

∠1+ ∠2=180-90

∠1+ ∠2=90            ——————-1

Similarly, in ∆DMC,

∠3+ ∠4=90°      _______2

∴ from 1 and 2

∠1+ ∠2=∠2+ ∠3

∠1=∠3

From 2 and 3

∠3+ ∠4=∠2+ ∠3

∠4=∠2

Now in ∆BMD and ∆DMC,

∠1=∠3

∠2=∠4

∴∆BMD~∆DMC           ——-(AA similarity criteria)

$\frac{BM}{DM} = \frac{MD}{MC}$

$\frac{DN}{DM} = \frac{MD}{MC}$

DM²=DN*MC

II) proceeding as in (i), we can prove that ∆BND~∆DNA

$\frac{BN}{DN} = \frac{ND}{NA}$

$\frac{DM}{DN} = \frac{ND}{NA}$      [BN=NA]

DN² = MN*MC

问题3。在图中，ABC是一个三角形，其中∠ABC> 90°，并且产生AD⊥CB。证明AC ² = AB ² + BC ² + 2BC.BD。

Given: In ∆ABC, ∠ABC > 90° and AD⊥CB provided.

To prove: (AC)²= AB²+BC²+ 2 BC.BD

Proof: In ∆ADB by Pythagoras theorem,

AC²=AD²+AB²

(AC)²= (AD)²+(DB)²        ———1

In ∆ADC by Pythagoras theorem,

(AC)²= (AD)²+(DC)²        ———2

AC²= AD²+DB²         ———1

AC²= AD²+(DB)²+2*DB*BC  

AC²= AD²+(DB)²+BC²+2*DB*BC  

问题4。在图中，ABC是一个三角形，其中∠ABC<90°，AD ADBC。证明AC ² = AB ² + BC ² –2BC.BD。

解决方案：

Given: In ∆ABC, ∠ABC < 90° and AD ⊥ BC.

To prove: AC² = AB² + BC² – 2*BC.BD

Proof: In triangle ADB, by  Pythagoras theorem

AB²=AD² +BD²       ———1

In triangle ADC, by  Pythagoras theorem

AC²=AD²+DC²       ———2

AC²=AD²+ BC² +BD²

AC²=AD²+ BC² +BD² – 2*BC*BD

AC²=AB²+ BD²+BC² – 2*BC*BD

AC² = AB² + BC² – 2BC.BD

问题5。在图中，AD是三角形ABC和AM⊥BC的中值。证明：

(i)AC ² = AD ² + BC.DM +(BC / 2) ²

(ii)AB ² = AD ² – BC.DM +(BC / 2) ²

(iii)AC ² + AB ² = 2 AD ² + 1/2 BC ²

Given: In ∆ABC, AD is a median, AM ⊥ BC.  

Prove that:

(i) AC² = AD² + BC.DM + (BC/2)²

(ii) AB² = AD² – BC.DM + (BC/2)²

(iii) AC² + AB² = 2 AD² + 1/2 BC²

Proof: i) In obtuse ∆ADC,∠D>90°

AC² = AD² + DC²+2DC.DM

AC² = AD² +(BC/2)²+2(BC/2).DM

AC² = AD² +(BC/2)2+BC.DM

Or   AC² = AD² +BC.DM+(BC/2)²

II)In acute triangle ABD,

∠D>90°  

AB² = AD² + BD²+2BD.DM

AB² = AD² +(BC/2)²-2(BC/2).DM

AB² = AD² +(BC/2)²-BC.DM

Or  

AB² = AD² -BC.DM +(BC/2)²

iii) Adding i and ii

AC² +AB² =AD²+BC.DM+(BC/2)² +AD² -BC.DM +(BC/2)²

             = 2 AD²+BC/4² +BC/4²

             = 2AD²+1/2.BC/4²

             = 2AD²+ BC²

问题6：证明平行四边形对角线的平方和等于其平行边线的平方和。

解决方案：

Given: ||gm ABCD and diagonals BD and AC

To prove: AB²+BC²+CD²+AD²=AC²+BD²

Proof: Diagonals of ||gm bisects each other at midpoint  

Solution: In ∆ABC, BO is median

∴AB²+BC²=2BO²+1/2AC²              ———1

In ∆ADC, DO is median

∴AD²+DC² = 2DO²+1/2AC²              ———2

Adding 1 and 2

AB²+BC²+AD²+DC²=2BO²+1/2AC²+2DO²+1/2AC²  

             =2.(1/2.BD)² +1/2AC²+2(1/2BD)²+1/2AC²

             =2.BD/4² +1/4AC²+2.BD/4²+1/2AC²

             =BD²/2+1/2AC² +BD²/2+1/2AC²

             =BD²+AC²

问题7。在图中，两个和弦AB和CD在P点处相交。证明：

(i)ΔAPC〜ΔDPB

(ii)AP.PB = CP.DP

解决方案：

Given: chords AB and CD intersect each other at the point P.

To prove: (i) Δ APC ~ Δ DPB           (ii) AP.PB = CP.DP

Proof: In ∆ABC and ∆DPB,

∠1=∠2            ——-[vertically opposite angle]

∠A=∠D           ——-[angle in the same segment of circle are equal]

i) ∴∆APC~∆DPB    (AA similarity criteria)

ii) $\frac{AP}{DB} = \frac{CP}{PB}$

AP*PB=CP*DP

问题8。在图中，圆的两个和弦AB和CD在圆外的P点(产生时)彼此相交。证明

(i)ΔPAC〜ΔPDB

(ii)PA.PB = PC.PD

解决方案：

i) In ∆PAC and ∆PDB,

∠P=∠P           ———[common]

∠1+∠2=180° ——–[linear pair angle]

∠2=180°+∠1

∠2+∠3=180° ——[sum of opposite angle of a cyclic quadrilateral is 180°]

∠3=180°-180°+∠1

∠3=∠1                   ————2

∠1=∠3 ———–[form 2]

∴∆APC~∆PDB        ———[AA similarity criteria]

ii) $\frac{PA}{PD} = \frac{PC}{PB}$

PA*PD = PC*PB

问题9。在图中，D是ΔABC的BC边上的一个点，使得$\frac{BD}{CD} = \frac{AB}{AC}$ 。证明AD是∠BAC的平分线。

解决方案：

Construction: Produce BA to E such that AE=AC join CE.

Proof: In ∆AEC, 

Since AE=AC

∴∠1=∠2             ——[Angle opposite to equal sides of ∆are=]

$\frac{BD}{CD} = \frac{AB}{AC}$  

$\frac{BD}{CD} = \frac{AB}{AE}$        ———–(BY CONSTRUCTION)

By converse of BPT

AD||EC

And AC is a transversal then

∠3=∠1       ——-[corresponding angles]

∠4=∠2       ——-[alternate interior angles]

But ∠1=∠2      ——–[by construction]

∴AD bisects ∠BAC internally

问题10：纳齐马在一条小溪中钓鱼。她的钓鱼竿的顶端在水面以上1.8 m，而字符串末端的蝇蝇则位于距离钓鱼竿顶端正下方3.6 m处和2.4 m处的水面上。假设她的字符串(从杆的尖端到苍蝇)绷紧，那么她有多少字符串(见图)?如果她以每秒5厘米的速度拉字符串，那么12秒钟后苍蝇到她的水平距离将是多少?

解决方案：

In ∆ABC,

AC is string

By Pythagoras theorem,

AC²=AB²+BC²

AC²=AB²+BC²

AC²=(1.8)²+(2.4)²

(AC)²=3.24+5.76

AC=√9

AC=3m

∴Length of string she have out=3m  

=5cm/sec*12sec

=60cm

=60/100 =0.6m  

AP=AC-0.6

=3-06

=2.4 m

Now in ∆ABP by Pythagoras theorem,

(AP)²=(AB)²+(PB)²

(2.4)²=(1.8)²+(PB)²

5.76=3.24 + (PB)²

5.76=3.24 + (PB)²

2.52=(PB)²

√2.52= PB

1.59=PB  

PD=PB+BD

PD=1.59+1.2

PD=2.79m  ———(approx)

Hence, the horizontal distance of the fly from nazima of the 12 sec in 2.79m.