Question

1) The two chord PQ and RS of a circle intersects each other at the point X within the circle. By joining P, S and R,Q, let us, ∆PXS and ∆RSQ are similar. From this, let us prove that PX:XQ = RX:XS. ​

Answer 1

Answer:

Given, chords RP=RQ

In △PSQ and △PSR

PQ=PR (given)

∠RPS=∠QPS (given)

PS=PS (common)

△PSQ≅△PSR (by SAS)

⇒RS=QS∠PSR=∠PSQ

But,

∠PSR+∠PSQ=180o2∠PSR=180o∠PSQ=∠PSR=90o

then, RS=QS and ∠PSR=90o

PS is the perpendicular bisector of chord RQ

PS passes through center of circle.

Answer 2

Answer:

$\sf\green{\boxed{Given :-}}$

↦ The two chord PQ and RS of a circle intersects each other at the point X within the circle. By joining P,S and R,Q, let us, ∆PXS and ∆RSQ are similar.

$\sf\green{\boxed{Construction :-}}$

↦ S,P and Q,R are joined.

$\sf\green{\boxed{Proof :-}}$

↦ Now , ∠SPQ and ∠SRQ are same circular angle SQ

➠ ∠ SPQ = ∠ SRQ i.e., ∠SPX = ∠XRQ

Again, ∠PSR and ∠ PQR are same circular angle PR

➠ ∠ PSR = ∠ PQR

i.e., ∠PSX = ∠ XQR

↦ Now, in ∆ PSX and ∆ROX, ∠ SPX =∠ XPQ

∠ PSX =∠ XQR

∴ ∆PSX and ∆RQX are equiangular. (Proved)

Now,

⇒ $\sf\dfrac{PX}{RX}$ = $\sf\dfrac{XS}{XQ}$

∴ PX : XQ = RX : XS (Proved)