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I try to do this problem:

It is given a rhombus $ABCD$ with sidelength $a$. On the line $AC$ are chosen the points $M$ and $N$ in such a way that $C$ lies between $A$ and $N$ and $MA\cdot NC= a^{2}$. We denote with $P$ the intersection point of $MD,BC$ and $Q$ is the intersection point of $ND,AB$. Prove that $D$ is the incenter of the triangle $PQB$.

I do this figure rhombus

By the sinus theorem applied to the triangles $MAD$ and $NCD$ implies $\alpha+\beta=\theta$ and these triangles are similar.

Then I've drawn the circumcircle to the triangle $BPQ$.

I wonder if the points $M$ and $N$ are on this circumcircle, if they are, the inscribed angles $NMP$ and $NQP$ are equal and the prove is done.

I appreciate any help, thanks.

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1 Answer 1

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Continuing your work:

Since $\angle AMD=\angle AQD$, quadrilateral $AMQD$ is cyclic. Therefore $\angle MQA=\beta$ and $\angle MQD=\alpha+\beta$.

Similarly, we can prove that $\angle NPD=\alpha+\beta$.

Therefore $\angle MQN=\angle MPN\implies MQPN\text{ is cyclic.}$

.. and the result follows.

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