let $I$ be the midpoint of $BC$ we have $IEC \sim CEA\implies \angle MEC =\angle NEA $ more $\angle QEC = \angle QEM +\angle MEC=90^\circ -\angle ENM+\angle MEC=90^\circ -\angle NEA+\angle MEC = 90^\circ -\angle MEC+\angle MEC=90^\circ.$ idem $\angle PEB=90^\circ $ therefore $\angle PEQ =\angle A$ RH HAS

I am having trouble understanding how $\angle PEB=90^{\circ} \implies \angle PEQ=\angle A$.

Plops wrote: I am having trouble understanding how $\angle PEB=90^{\circ} \implies \angle PEQ=\angle A$. $\angle PEB=90^{\circ}$; $\angle QEC=90^{\circ}$ So $\angle BAC = 180^{\circ} - \angle BEC = 180^{\circ} - (\angle PEB + \angle QEC - \angle PEQ) = \angle PEQ$

Good problem in Baltic Way style. Solved with @blastoor We begin with the following crucial claim. Claim: $BE$ is tangent to $\odot(KLE)$ and $CE$ is tangent to $\odot(NME)$. Proof: Note that $\triangle BDE \sim ABE $. This implies that: $$ \frac{BD}{BE} = \frac{AB}{AE} \implies \frac{BL}{BE} = \frac{AK}{AE} $$Consequently we have that $\triangle ABE \sim \triangle AKE$, which gives us that $\angle BEL = \angle KEA = \angle LKE$. This proves that $BE$ is tangent to $\odot(KLE)$. Similarly we can prove that $CE$ is tangent to $\odot(NME)$. From claim follows that $\angle BEP = \angle QEC = 90^{\circ} \implies \angle BEQ = \angle PEC$. Claim: $\angle BEQ = 90^{\circ} - \angle BAC = \angle PEC$ Proof: Denote $x= \angle AEN = \angle ENM = \angle MEC$. Note that: $$ \angle BEQ = \angle BEM - \angle QEM = \angle BEM - 90^{\circ}+x $$But note that $BEM = 180^{\circ} - \angle BAC - \angle MEC = 180^{\circ} - \angle BAC -x$. Thereby: $$ \angle BEQ =\angle BEM -90^{\circ}+x = 180^{\circ} - \angle BAC -x -90^{\circ}+x=90^{\circ} - \angle BAC $$ Now to finish the problem we are left to compute $\angle PEQ$. This is easy since: $$ \angle PEQ = \angle BEM - \angle BEQ - \angle PEC = 180^{\circ} - \angle BAC -90^{\circ}+\angle BAC -90^{\circ}+\angle BAC =\angle BAC$$as desired.

Kimchiks926 wrote: Good problem in Baltic Way style. Solved with @blastoor We begin with the following crucial claim. Claim: $BE$ is tangent to $\odot(KLE)$ and $CE$ is tangent to $\odot(NME)$. Proof: Note that $\triangle BDE \sim ABE $. This implies that: $$ \frac{BD}{BE} = \frac{AB}{AE} \implies \frac{BL}{BE} = \frac{AK}{AE} $$Consequently we have that $\triangle ABE \sim \triangle AKE$, which gives us that $\angle BEL = \angle KEA = \angle LKE$. This proves that $BE$ is tangent to $\odot(KLE)$. Similarly we can prove that $CE$ is tangent to $\odot(NME)$. From claim follows that $\angle BEP = \angle QEC = 90^{\circ} \implies \angle BEQ = \angle PEC$. Claim: $\angle BEQ = 90^{\circ} - \angle BAC = \angle PEC$ Proof: Denote $x= \angle AEN = \angle ENM = \angle MEC$. Note that: $$ \angle BEQ = \angle BEM - \angle QEM = \angle BEM - 90^{\circ}+x $$But note that $BEM = 180^{\circ} - \angle BAC - \angle MEC = 180^{\circ} - \angle BAC -x$. Thereby: $$ \angle BEQ =\angle BEM -90^{\circ}+x = 180^{\circ} - \angle BAC -x -90^{\circ}+x=90^{\circ} - \angle BAC $$ Now to finish the problem we are left to compute $\angle PEQ$. This is easy since: $$ \angle PEQ = \angle BEM - \angle BEQ - \angle PEC = 180^{\circ} - \angle BAC -90^{\circ}+\angle BAC -90^{\circ}+\angle BAC =\angle BAC$$as desired. Your solution is very nice and beautiful but I think this is slightly quicker when proving $\angle BEQ = 90^{\circ} - \angle BAC = \angle PEC$. After proving that $\angle BEP$ = $90^{\circ}$ = $\angle QEC$ one can quickly deduce that $$180^{\circ} - \angle QEP = 90^{\circ} + 90^{\circ} -\angle QEP = \angle BEP + \angle QEC - \angle QEP = \angle BEC$$

Here is another solution which is not as elegant but very straightforward since it does not require any clever observation: