A circle is inscribed in trapezoid $ABCD(BC \parallel AD)$. The circle is tangent to the sides of $AB$ and $CD$ at $K$ and $L$, respectively.

Question

A circle is inscribed in trapezoid $ABCD(BC \parallel AD)$. The circle is tangent to the sides of $AB$ and $CD$ at $K$ and $L$, respectively, and to bases $AD$ and $BC$ at $M$ and $N$, respectively. Prove that $AK\cdot KB = CL\cdot CD$.

I'm reading the solution to this problem but I'm unable to verify how the Author(Prasolov) deduced he final answer.

He first noted that since $ABCD$ is a tangential trapezoid, then $\measuredangle CBA + \measuredangle BAD = 180^o$

And since $O$ is the incentre of $ABCD$ then $\measuredangle OBA + \measuredangle OAB = 90^o$ since $OA$ and $OB$ are angle bisectors. He went on to assert that triangle $AKO$ is similar to triangle $OKB$

But I can't justify this assertion, I need someone to complete the solution for me. Thanks.

Answer 1

I think you mean we need to prove that $AK\cdot KB=CL\cdot LD.$

Since $\measuredangle AOB=\measuredangle COD=90^{\circ},$ we obtain:

$$AK\cdot KB=OK^2=OL^2=CL\cdot LD$$

I used the following fact.

Let $\measuredangle ACB=90^{\circ}$ and $CD$ be an altitude of $\Delta ABC$. Prove that: $$CD^2=AD\cdot DB.$$

It follows immediately from $\Delta ACD\sim\Delta CBD$.