Question

Two circles with centre a and b intersect each other at p and q and m is the mid point of pq. Say if the following are true
AM perpendicular to PQ
BM perpendicular to PQ
A,M and B are collinear​

Answer 1

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✤ Required Answer:

► GiveN:

• Two circles of centre A and B intersect each other at P and Q.
• M is the midpoint of PQ.

► To do:

True or false✓

• AM is perpendicular to PQ.
• BM is perpendicular to PQ.
• A, M and B are collinear.

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✤ How to solve?

Let's try to determine true or false in the above statements by using congurence of triangles.

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✤ Solution:

1) AM perpendicular to PQ

In △AMP and △AMQ

• AP = AQ (Radius of same circle)
• AM = common
• PM = QM (Perpendicular drawn from the centre to the chord bisects it)

So, △AMP ≅ △AMQ (By SSS congurence)

➝ ∠AMP = ∠AMQ (CPCT)

We know,

• PQ is a straight line, then

➝ ∠AMP + ∠AMQ = 180°

➝ 2∠AMP = 180°

➝ ∠AMP = 90°

And, ∠AMQ = 90°

So, the statement is:

$\large{ \boxed{ \bf{ \red{True}}}}$

2) BM perpendicular to PQ

In △BMP and △BMQ

BP = BQ (Radius of same circle)

BM = common

PM = QM (Perpendicular drawn from the centre to the chord bisects it)

So, △BMP ≅ △BMQ (By SSS congurence)

➝ ∠BMP = ∠BMQ (CPCT)

We know,

PQ is a straight line, then

➝ ∠BMP + ∠BMQ = 180°

➝ 2∠BMP = 180°

➝ ∠BMP = 90°

And, ∠BMQ = 90°

So, the statement is:

$\large{ \boxed{ \bf{ \red{True}}}}$

3) A,M and B are collinear

From (1) and (2), we got

• ∠AMP = 90° -------(1)
• ∠BMP = 90° --------(2)

By adding (1) and (2),

➝ ∠AMP + ∠BMP = 180°

➝ ∠AMB = 180°

Hence,

• A, M and B lies on the same straight line and are collinear.

So, the statement is:

$\large{ \boxed{ \bf{ \red{True}}}}$

Hence, solved !!

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