Question

In the given figure, two circles intersect at two points A and B. XY is a tangent at the point P. Prove that CD is parallel to the tangent XY

Answer 1

If two circles intersect at two points A and B and XY is a tangent at the point P then it is proved that CD is parallel to the tangent XY.

Step-by-step explanation:

It is given that,

Two circle intersect at points A and B.

XY is a tangent at point P

From the figure attached below, we can see that from point P of one of the two circles, two segments PAC and PBD are drawn which intersect another circle at point C and D.

Also, join point A and B.

We know that according to the alternate segment theorem the angle between a chord and a tangent through one of the endpoints of the chord is equal to the angle in the alternate segment.

Here XY is a tangent

∴ ∠APX = ∠ABP ……….. (i)

Now, from the figure again we can see that all the vertices of ABCD lie inside the circle so it is a cyclic quadrilateral.

So, by the theorem that the sum of the opposite angles of a cyclic quadrilateral is 180°, we get

∠ABD + ∠ACD = 180° …… (ii)

Also, ∠ABD + ∠ABP = 180° …….. [Linear Pair] ….. (iii)

Therefore, from (ii) & (iii), we get

∠ACD = ∠ABP …….. (iv)

From (i) and (iv), we get,

∠ACD = ∠APX

Since we know that when two lines are parallel to each other and the angles lying on the opposite sides of the transverse line i.e., the alternate angles are equal to each other, so we can conclude that

XY // CD  

Hence Proved.

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Also View:

The given figure shows two parallel tangents xy and x'y' at the points a and b respectively to the circle with cetre o. another tangent cd is drawn paralel to ab at the point p. show that a circle with p as the centre and op as radius will pass through c and d .

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Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.

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Answer 2

Answer:

Answer is in above

Step-by-step explanation:

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