Question

Triangle ABC is inscribed in a circle with center O'. A circle with center O is inscribed in triangle ABC. AO is drawn, and extended to intersect the larger circle in D. Then we must have.

• A. CD $=$ BD $=$ O'D
• B. AO $=$ CO $=$ OD
• C. CD $=$ CO $=$ BD
• D. CD $=$ OD $=$ BD
• E. O'B $=$ O'C $=$ OD

Answer 1

The correct option is D CD $=$ OD $=$ BD

We will prove that $\Delta DOB$ and $\Delta COD$ is isosceles,

meaning that $CD=OD=BD$.

Let $\angle A=2\alpha$ and $\angle B=2\beta$.

Since the incentre of a triangle is the intersection of its angle bisectors,

$\angle OAB=\alpha$ and $\angle ABO=\beta$.

Hence $\angle DOB=\alpha +\beta$.

Since quadrilateral $ABCD$ is cyclic,

$\angle CAD=\alpha =\angle CBD$.

So $\angle OBD=\angle OBC+\angle CBD=\alpha +\beta =\angle DOB$.

This means that $\Delta DOB$ is isosceles, and hence $BD=OD$.

Now let $\angle C=2\gamma$ which means $angleACO=COD=\gamma$.

Since$ABCD$ is cyclic, $\angle DAB=\alpha =\angle DCB.$ Also, $\angle DAC\alpha$

so $\angle DOC=\alpha +\gamma$.

Thus, $\angle OCD=\angle OCB+\angle BCD=\gamma +\alpha =\angle DOC$

which means $\Delta COD$ is isosceles, and hence $CD=OD=BD$.