Mathematics
ΔABC is an isosceles triangle in which AB = AC, circumscribed about a circle. Prove that the base is bisected by the point of contact.
Answer
We know that,
Tangents from exterior point are equal in length.
We have,
AR = AP, BQ = BP and CQ = CR
Now, AB = AC
⇒ AP + PB = AR + RC
⇒ AR + PB = AR + RC [∵ AR = AP]
⇒ PB = RC
⇒ BQ = CQ.
It means BC is bisected at point Q.
Hence, proved that the base is bisected by the point of contact.
Related Questions
Show that the line segment joining the points of contact of two parallel tangents passes through the centre.
In the given figure, PQ is a transverse common tangent to two circles with centres A and B and of radii 5 cm and 3 cm respectively. If PQ intersects AB at C such that CP = 12 cm, calculate AB.
In the given figure, quadrilateral ABCD is circumscribed and AD ⟂ AB. If the radius of the incircle is 10 cm, find the value of x.
In the given figure, a circle is inscribed in quadrilateral ABCD. If BC = 38 cm, BQ = 27 cm, DC = 25 cm and AD ⟂ DC, find the radius of the circle.
