Exercise 14 D
Question 1
Ans: In a circle with center O, from a paint P outside of it, PA and PB are the tangents to the circle
At a point m on the circle another tangent is drawn which intersects AP at K and at N.
To prove: KN = AK+BN
Proof: From K, KA and KM are tangents to the circle
So KA=KM ........(i)
Similarly from $N, N M$ and $N B$ tangents are drawn
So $N B=m N$
Adding i and ii
$\begin{aligned}KM+M N &=k A+N B \\k N &=A K+B N\end{aligned}$
Question 2
Ans: Two concentric circles with center O .
$A$ chord $A B$ is drawn to the greater circle which touches the smaller circle at $P$.
To prove: $A B$ is bisected at $P$ i.e. $A P=P B$ construction: Join DP, OA and OB.
(IMAGE TO BE ADDED)
Proof: if $O P$ is the radiues and $A B$ is the tangent
So OP $\perp A B$
In right $\triangle O A P$ and $\triangle O B P$
HyP. $O A=O B$
Side $O P=O P$
So $\triangle O A P \equiv \triangle O B P$
So $A P=P B$
Hence $P$ bisects $A B$ Hence Proved
Question 3
Ans: A circle with center O, from a point P outside of it, two tangents PT and PS are drawn TS is
Joined which subtends
ㄥTOS at the centre
To prove: $\angle T P S+\angle T O S=180^{\circ}$
Construction: Join OP.
(IMAGE TO BE ADDED)
Proof: In right angle $\Delta O T P$
$\angle O T P=90^{\circ}$
So $\angle T O P+\angle T P O=90^{\circ}$.........(i)
Similarly in right $\triangle O S P$
$\angle S O P+\angle S P O=90^{\circ}$.......(ii)
Adding (i) and (ii)
$\begin{aligned}&\angle T O P+\angle S O P+\angle T P O+\angle S P O=90^{\circ}+90^{\circ} \\&\Rightarrow \angle T P S+\angle T O S=180^{\circ}\end{aligned}$
So LTPS and ㄥTOS are supplementary hence proved
Question 4
Ans: In the figure a circle touches the side BC of a $\triangle A B C$ externally and AB and AC on producing at Y and Z
To prove: $A Y=\frac{1}{2}(A B+B C+C A)$
Proof: if from $A, A Y$ and $A Z$ are tangents to the circle
So $A Y=A Z$
similarly from $B, B y$ and $B x$ are tangents So $B y=B x$
and from CX and CZ are tangents
So C Z=C X
Now $A Y=A B+B Y=A B+B \times$
$\begin{aligned}&A Z=A C+C Z=A C+C X \\&\text { So } A Y+A Z=A B+B \times+C \times+A C \\&\Rightarrow A Y+A Y=A B+B C+C A \\&\Rightarrow 2 A Y=A B+B C+C A \\&\Rightarrow A Y=\frac{1}{2}(A B+B C+C A)\end{aligned}$
Hence proved
Question 5
Ans: In circle touches the sides BC , CA and AB at D,E and F respectively of the $\triangle A B C$
To prove:
$\begin{aligned} A F &+B D+C E=A E+C D+B F \\ &=\frac{1}{2}(A B+B C+C A) \end{aligned}$
Proof: from $A, A F$ and $A E$ are the tangents to the circle
So $A F=A E$.......(i)
Similarly From B, BD and BF are the tangents
So $B D=B F$......(ii)
and From C, CE and $C D$ are the tangents
So $C E=C D$.........(iii)
Adding (i) (ii) and (iii) we get,
$A F+B D+C E=A E+B F+C D$
Adding to both sides
A F+B D+C E
$2 A F+2 B D+2 C E=A E+B F+C D+A F+B D+C E$
$2(A F+B D+C E)=A E+C E+B D+C D+A F+B F$
$\Rightarrow 2(A F+B D+C E)=A C+B C+A B$
$\Rightarrow A F+B D+C D=\frac{1}{2}(A B+B C+C A)$ Hence proved
Question 6
Ans: Two circles touch each other externally at $T$.
$P T$ is their common tangent and from $P$, tangent $P A$ and $P B$ are drawn to the two circles
To prove: $P A=P B$
proof: if from P,PA and PT are the tangents to the bigger circle
So $\quad P A=P T$...........(i)
Similarly from $P, P T$ and $P B$ are the tangents drawn to the smaller circle
So $P T=P B$........(ii)
from (i) and (ii)
P A=P B Hence proved
Question 8
Ans: Two circles touches each other at A externally. Through A, a common tangent is drawn. A point $P$ is taken on the tangents.
(IMAGE TO BE ADDED)
To prove : $P B=P C$
-Proof: Through $P_{1}$ two tangents $P A$ and $P B$ and drawn to the first circle So P A=P B......... (i)
Again through P, AB and $P C$ are the tangents drawn to the second circle
So $\quad P A=P C$
From (i) and (ii)
PB= PC Hence proved
Question 9
Ans: In the circle with center 0 , from a point $p$ outside of it PA and PB are tangents drawn $O P, O A, O B$ and $A B$ are joined
To prove:
(i) $\angle A O P=\angle B O P$
(ii) OP is the perpendicular bisector of AB
proof:
(i) In right $\triangle O A P$ and $\triangle O B P$.
Hyp OP =OP
side $O A=O B$
So $\triangle O A P \cong \triangle O B P$
So $\angle A O P=\angle B O P$
and $\angle A P O=\angle B P O$
Now in $\triangle A M P$ and $\triangle B M P$
$MP=MP$
$P A=P B$
$\angle A P M=\angle B P M$
So $\triangle A M P \cong \triangle B M P$
So $A M=B M$
$\angle A M P=B M P$
But $\angle A M P+\angle B M P=180^{\circ}$
So $\angle A M P=\angle B M P=90^{\circ}$
Hence OP is the perpendicular bisects of AB
Question 10
Ans: In a circle with center O, AB is the chord
Through A and B tangents PQ and PR are drawn which meet each other at P
(IMAGE TO BE ADDED)
To prove: $\angle P A B=\angle P B A$
construction: Join $O A, O B$ and $O P$
Proof: In $\triangle O A P$ and $\triangle O B P$,
$O P=O P$
$O A=O B$
$P A=P B$
So $\triangle O A P \cong \triangle O B P$
So $\angle A P O=\angle B P O$
Now in $\triangle P A M$ and $\triangle P B M$,
$\begin{aligned}&P M=P M \\&P A=P B\end{aligned}$
$\angle A P M=\angle B P M ( A P O=\angle B P O)$
So $\triangle P A M \cong \triangle P B M$
So $\angle P A M=\angle P B M$
$\Rightarrow \angle P A B=\angle P B A$
Question 11
Ans: Two circles with centers O and O' touch each other externally at C
$P Q$ and $R S$ are common tangents and a third common tangent from $C$ is drawn which intersecting $P Q$ and $R S$ and $L$ and M
To prove: LM bisects PQ and RS
Proof: From L, LP and LC are the tangents to the circle
So LP = LC .........(i)
Similarly from and L, LQ and LC are two tangents
SO LC =L Q$
from (i) and (ii)
LP = LC =LQ
So L is the mid point of PQ
Similarly we can prove that
$m$ is the mid point of RS
Hence $L m$ bisects the tangents $P Q$ and RS
Question 12
Ans: A circle with center C . $P A$ and $P B$ are two tangents drawn from $P$ to the circle $C A$ is joined.
To prove: $\angle A P B=2 \angle C A B$
Proof: In $\triangle A P B$,
$A P=B P$
So $\angle P A B=\angle P B A$
But $\angle P A B+\angle P B A+\angle A P B=180^{\circ}$
$\begin{aligned}\Rightarrow & \angle P A B+\angle P A B+\angle A P B=180^{\circ} \\\Rightarrow & \angle A P B=180^{\circ}-2 \angle P A B \\& \text { But } \angle P A B=\angle C A P-\angle C A B \\&=90^{\circ}-\angle C A B \\\text { So } \angle A P B=180^{\circ}-2\left(90^{\circ}-\angle C A B\right) \\\Rightarrow & \angle A P B=180^{\circ}-180^{\circ}+2 \angle C A B \\\Rightarrow & \angle A P B=2 \angle C A B\end{aligned}$
Hence proved
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