Question

In figure 3.76, seg PQ is a diameter of a
circle with centre O. R is any point on the
circle
seg Rs I seg PQ.
Prove that, SR is the geometric mean
of PS and SQ.
[That is, SR' = PS X SQ]

Hint: Write the proof with the help of the following steps.
(1) Draw ray RS. It intersects the circle at T.
(2) Show that RS = TS.
(3) Write a result using theorem of intersection of chords inside the circle.
(4) Using RS = TS complete the proof.​

Answer 1

In figure 3.76, seg PQ is a diameter of a  circle with centre O. R is any point on the  circle  seg Rs I seg PQ.

Steps of construction:

Join OT and OR.

Hence we have obtained two right angles triangles Δ OSR and Δ OST.

In Δ OSR and Δ OST

OS = OS (common side)

∠ S = ∠ S = 90° (common angle)

OR = OT  (radii of same circle)

Therefore, from SAS criteria theorem, we have,

Δ OSR  ≅ Δ OST

⇒ RS = TS ..........(1)

The theorem of intersection of chords inside the circle states that, "When two chords intersect each other inside a circle, the product of their segments are equal to each other."

Therefore we have,

PS * SQ = RS * TS

as, RS = TS   (from (1))

PS × SQ = RS × RS

∴ (SR)² = PS × SQ

Hence it is proved that, SR is the geometric mean  of PS and SQ.