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Given a parallelogram $ABCD$ . Points $M$ and $N$ are respectively the midpoints of $BC$ and $CD$ . Lengths $AM$ and $AN$ intersecting diagonal $BD$ consecutive points $P$ and $Q$. Prove that triangles $AQB$ I $NQD$ are similar and find the coefficient of similarity . Prove that the length of $BP$ , $PQ$ and $QD$ equal length .
My work: If triangle $AQB$ I $NQD$ are similar then we know: $$\frac{AQ}{NQ}=\frac{AB}{ND}=\frac{QB}{QD}=k$$ $$k=\frac{AB}{ND}=\frac{CD}{ND}=\frac{2ND}{ND}=2$$ By the diagonal $BD$ we have 1 equal angle $\angle QBA=\angle QDN $

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  • $\begingroup$ A diagram of the situation would be helpful. $\endgroup$ – String Apr 26 '16 at 11:04
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If triangle $AQB$ I $NQD$ are similar then

We have to prove that $\triangle{AQB}$ and $NQD$ are similar, so you cannot start with that.


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Since $CD$ is parallel to $BA$, we have $$\angle{QAB}=\angle{QND}\quad\text{and}\quad \angle{QBA}=\angle{QDN}$$ from which the claim follows with $QB:QD=AB:ND=2:1$.

Similarly, we can prove that $\triangle{APD}$ and $\triangle{MPB}$ are similar with $DP:BP=AD:MB=2:1$.

Thus, we can write $$BP=a,\quad DP=2a,\quad DQ=b,\quad QB=2b$$ From $BD=3a=3b\Rightarrow a=b$, we have $BP=PQ=QD$.

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