Question

Two circles with center A and B intersect each other at points P and Q. Prove that the line
joining the centers A and B bisects the common chord PQ at right angles.​

please answer in full step.

Answer 1

Hope it helps!! Mark this answer as brainliest if u found it useful and follow me for quick and accurate answers...

Answer 2

Answer:

In ΔAXBandΔAYB,

AX=AY (Radii of the same circle)

BX=BY (Radii of the same circle)

AB=AB (Common)

$$∴ ΔAXB ≅ ΔAYB$$ (SSS criterion)

$$⇒ ∠4 = ∠5$$ (CPCT) ---(1)

In AOXandΔAOY,

$$∠4 = ∠5$$ (From 1)

AX=AY (Radii of the same circle)

AO=AO (Common)

$$∴ ΔAOX ≅ ΔAOY$$ (SAS criterion)

$$⇒∠1 = ∠2$$ (CPCT) 

But$$ ∠1 + ∠2 = 180°$$ (Linear Pair Axiom)

$$2∠1 = 180°$$

$$⇒ ∠1 = 90°$$

$$⇒ ∠2 = 90°$$

Now,$$ CD║XY$$

Therefore$$ ∠1 = ∠3$$ (Alternate interior angles)

Or $$∠3 = 90°$$ 

Since $$AM⊥PQ and AM passes through B (The center), 

$$⇒PM=PQ(A$$ perpendicular from the center to a chord bisects the chord)

Hence Proved