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# Given $P$ is a point of intersection of two circles with centers $C$ and $D$ . If the straight line $APB$ is parallel to $CD$ , thenA. $AB = CD$ B. $AB = \dfrac{1}{2}CD$C. $3AB = CD$D. $AB = 2CD$

Last updated date: 20th Jun 2024
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Hint: Here we will first draw two circles and then we will draw the point of intersection of two circles and draw everything given in the question. We will consider all the conditions given in the questions and then we will consider two triangles in the circle and we will make them congruent. From there, we will get the required relation.

Now, we will draw two circles with their point of intersection as $P$ and it is given that the straight line $APB$ is parallel to $CD$ . So, we will draw $CC' \bot AB$ and $DD' \bot AB$.
It is given that $AB$ is parallel to $CD$.
As we know $CC' \bot AB$ and $DD' \bot AB$.
So, we can say that
$CC' = DD'$
And hence
$C'D' = CD$ ………. $\left( 1 \right)$
Now, we will consider $\vartriangle AC'C$ and $\vartriangle PC'C$ , then we will make them congruent to each other.
In $\vartriangle AC'C$ and $\vartriangle PC'C$, we have
$PC = AC$ because they are the radii of a circle.
$\angle AC'C = \angle PC'C = 90^\circ$
$CC' = CC'$ because they are common sides of the triangles.
Thus, $\vartriangle AC'C \cong \vartriangle PC'C$ by SAS rule of congruence.
Therefore, by corresponding part of the congruent triangles, we get
$AC' = PC'$ ……….. $\left( 2 \right)$
Similarly, we will consider $\vartriangle PD'D$ and $\vartriangle BD'D$, then we will make them congruent to each other.
In $\vartriangle PD'D$ and $\vartriangle BD'D$, we have
$PD = BD$ because they are the radii of a circle.
$\angle PD'D = \angle BD'D = 90^\circ$
$DD' = DD'$ because they are common sides of the triangles.
Thus, $\vartriangle PD'D \cong \vartriangle BD'D$ by SAS rule of congruency.
Therefore, by corresponding part of the congruent triangles, we get
$PD' = BD'$ ……….. $\left( 3 \right)$
From the figure, we know that,
$AB = AC' + PC' + PD' + BD'$ ……… $\left( 4 \right)$
From equation $\left( 2 \right)$ and equation $\left( 3 \right)$, we have;
$AC' = PC'$
$PD' = BD'$
Now, we will substitute these values in equation $\left( 4 \right)$. Therefore, we get
$\Rightarrow AB = PC' + PC' + PD' + PD'$
On adding the like terms, we get
$\Rightarrow AB = 2PC' + 2PD'$
On further simplification, we get
$\Rightarrow AB = 2\left( {PC' + PD'} \right)$ ……… $\left( 5 \right)$
We know that $PC' + PD' = C'D'$. So, substituting this value in the above equation $\left( 5 \right)$, we get
$\Rightarrow AB = 2C'D'$ ………. $\left( 6 \right)$
We know from equation $\left( 1 \right)$ that $C'D' = CD$.
So, replacing $C'D'$ by $CD$ in equation $\left( 6 \right)$, we get
$\Rightarrow AB = 2CD$
Hence, the correct option is option D.

Note: Here we have made two triangles congruent to each other. A triangle is a two-dimensional geometric shape that has two sides. Any two objects are congruent only if they superimpose on each other. Two triangles are said to be congruent if all their three angles and three sides are equal. But it is not necessary to find all the six dimensions. Hence, the two triangles can be made congruent by knowing only three values out of six values.