thereom of tangents that are equalornot
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STATEMENT :- The lengths of tangents drawn from an external point to a circle are equal.
Given:- A circle with center O,P is a point lying outside the circle and PA and PB are two tangents to the circle from P.
To prove :- PA = PB
Proof:- join OA ,OB and OP
<OAP = <OBP ------------------->(Angle between radii and tangents)
In two right triangles
OA = OB ------------------------->(radii of same circle)
OP = OP ------------------------->(common)
Therefore, by R.H.S congruency axiom,
ΔOAP ≡ ΔOBP
⇒PA = PB -------------------------->(CPCT)
HENCE PROVED
Given:- A circle with center O,P is a point lying outside the circle and PA and PB are two tangents to the circle from P.
To prove :- PA = PB
Proof:- join OA ,OB and OP
<OAP = <OBP ------------------->(Angle between radii and tangents)
In two right triangles
OA = OB ------------------------->(radii of same circle)
OP = OP ------------------------->(common)
Therefore, by R.H.S congruency axiom,
ΔOAP ≡ ΔOBP
⇒PA = PB -------------------------->(CPCT)
HENCE PROVED
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STATEMENT :- The lengths of tangents drawn from an external point to a circle are equal.
Given:- A circle with center O,P is a point lying outside the circle and PA and PB are two tangents to the circle from P.
To prove :- PA = PB
Proof:- join OA ,OB and OP
<OAP = <OBP ------------------->(Angle between radii and tangents)
In two right triangles
OA = OB ------------------------->(radii of same circle)
OP = OP ------------------------->(common)
Therefore, by R.H.S congruency axiom,
ΔOAP ≡ ΔOBP
⇒PA = PB -------------------------->(CPCT)
HENCE PROVED
Given:- A circle with center O,P is a point lying outside the circle and PA and PB are two tangents to the circle from P.
To prove :- PA = PB
Proof:- join OA ,OB and OP
<OAP = <OBP ------------------->(Angle between radii and tangents)
In two right triangles
OA = OB ------------------------->(radii of same circle)
OP = OP ------------------------->(common)
Therefore, by R.H.S congruency axiom,
ΔOAP ≡ ΔOBP
⇒PA = PB -------------------------->(CPCT)
HENCE PROVED
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