Question

prove that tangents drawn to a circle from an external point or equal

Answer 1

Step-by-step explanation:

To prove PT=QT

Proof: Consider the triangle OPT and OQT.

OP=OQ

∠OPT=∠OQT=90∘

OT=OT (common side)

Hence by RHS the triangles are equal.

Hence PT=QT

Hence Proved.

Answer 2

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Question

prove that tangents drawn to a circle from an external point or equal



Solution
Let two tangent PT and QT are drawn to circle of centre O as shown in figure. 
Both the given tangents PT and QT touch to the circle at P and Q respectively. 

We have to proof : length of PT = length of QT 
Construction :- draw a line segment ,from centre O to external point T { touching point of two tangents } . 

Now ∆POT and ∆QOT 
We know, tangent makes right angle with radius of circle. 
Here, PO and QO are radii . So, ∠OPT = ∠OQT = 90° 
Now, it is clear that both the triangles ∆POT and QOT are right angled triangle.
nd a common hypotenuse OT of these [ as shown in figure ] 

Now, come to the concept , 
∆POT and ∆QOT 
∠OPT = OQT = 90° 
Common hypotenuse OT 
And OP = OQ [ OP and OQ are radii]
So, R - H - S rule of similarity 
∆POT ~ ∆QOT 
Hence, OP/OQ = PT/QT = OT/OT 
PT/QT = 1 
PT = QT [ hence proved]



Note = Figure in the attachment



Hope it helps