Question

abc is an isosceles triangle in which ab=ac circumscribed about a circle. show that bc is bisected at the point of contact

Answer 1

we know that the tangents drawn from an external point to a circle are equal in length
So...
AP=AQ......(1)
BP=BR......(2)
CQ=RC......(3)
Now, the given triangle is an isosceles, So  AB=AC
Subtract AP from both sides, we get
AB-AP = AC-AQ (Using (1) )
BP=CQ
=> BR=CQ  (Using (2) )
=> BR=CR  (Using (3) )
So, BR=CR ( shows that BC is bisected at the point of contact)

Answer 2

Step-by-step explanation:

we know that the tangents drawn from an external point to a circle are equal in length

So...

AP=AQ......(1)

BP=BR......(2)

CQ=RC......(3)

Now, the given triangle is an isosceles, So  AB=AC

Subtract AP from both sides, we get

AB-AP = AC-AQ (Using (1) )

BP=CQ

=> BR=CQ  (Using (2) )

=> BR=CR  (Using (3) )

So, BR=CR ( shows that BC is bisected at the point of contact)