You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Angle in a Semicircle Is 90 Degrees

In Fig. 8.64,...

Question

In Fig. 8.64, PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at M. Prove that PL+ML=PN+MN.

Open in App

Solution

PA = PB [TANGENTS DRAWN FROM AN EXTERNAL POINT TO A CIRCLE ARE EQUAL IN LENGTH]

PL + AL = PN + BN [P-L-A, AND P-N-B]

AL = ML & BN = MN [SAME REASON AS 1]

Thus, PL + ML = PN + MN [FROM 2 AND 3]

Hence, proved.

Suggest Corrections

Similar questions

Q. In the given figure,

P A

and

P B

are tangents from an external point

P

to a circle with center

O

L N

touches the circle at

M

. Prove that

P L + L M = P N + M N

Q. In fig., PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4cm. If PA

⊥

PB, then the length of each tangent is

P A

and

P B

are tangents to the circle with centre

O

from an external point

P

, touching the circle at

A

and

B

respectively. Show that the quadrilateral

A O B P

is cyclic.

Q. From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB.

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of $Δ$ PCD.

Join BYJU'S Learning Program

In Fig. 8.64, PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at M. Prove that PL+ML=PN+MN.

PA = PB [TANGENTS DRAWN FROM AN EXTERNAL POINT TO A CIRCLE ARE EQUAL IN LENGTH] PL + AL = PN + BN [P-L-A, AND P-N-B] AL = ML & BN = MN [SAME REASON AS 1] Thus, PL + ML = PN + MN [FROM 2 AND 3] Hence, proved.

PA = PB [TANGENTS DRAWN FROM AN EXTERNAL POINT TO A CIRCLE ARE EQUAL IN LENGTH]

PL + AL = PN + BN [P-L-A, AND P-N-B]

AL = ML & BN = MN [SAME REASON AS 1]

Thus, PL + ML = PN + MN [FROM 2 AND 3]

Hence, proved.