Prove that $\measuredangle AQP+\measuredangle NAP=90^o$

Question

Let a triangle $ABC$ be right at $A$, $AH$ be the altitude to the side $BC$. Let $M$ be an arbitrary point located in $AH$. Draw circle $B$ with the radius $BA$, circle $C$ with the radius $CA$. The circle $B$ and the circle $C$ respectively cut $CM, BM$ at $P$ and $Q$. Circle $(HPQ)$ cut $BC$ at $N$. Prove that $$\measuredangle AQP+\measuredangle NAP=90^o$$

Here is the GeoGebra file of this image.

Answer 1

The following works only for the particular case when the circle PHNQ passes through A too.

$\alpha + \gamma = \beta + \gamma$ (angles in the same segment)

$= \delta$ (exterior angles cyclic quad)

$= 90^0$!

figure-1

For the general case, I have to give up after a long period of trial. However, I would like to share some of the interesting findings that might be useful for those who are interested.

(1) Some colored pairs of lines are parallel. XY should be parallel to PB’ too but that remains to be proven.

(2) PP’A’M’ is cyclic.

(3) If Point T is the perpendicular point of XTY and APTP’, then the question is solved.

Answer 2

The solution is complicated (and I hope someone can provide a simpler one). It will be subdivided into several parts. In addition, lots of constructions are needed and therefore written descriptions will all be omitted. I hope that the picture can clearly show how the new or extended lines are drawn and the intersection points are located.

In particular, $C’R$ is produced to cut $AB’$ at $T$. By angles in the same segment, all green-marked angles are all equal to $c$. This means that $C’T // SB’, \measuredangle ARC’ = 90^{\circ}$ and $\measuredangle ATC’ =\measuredangle AB’S$.

The circum-circle $AC’T$ is created (with $C’T$ as diameter). It can be shown that $\measuredangle AC’T = \measuredangle ATC’ = 45^{\circ}$.

We then add a few more lines to obtain the following diagram (pls ignore the red and blue circles):-

To get the required, it is sufficient to show that $\measuredangle AXR = 90^{\circ}$. Before that, we need a lemma with the following assumptions:-

1. Let $A$ be one of the endpoints of the common chord ($AC’$) of two intersecting circles.

2. Let $\phi$ be the angle between the tangents (at the same point to the two circles) through $A$.

3. Through $A$, a line is drawn cutting the two circles at $J$ and $K$.

Conclusion:- $\measuredangle JC’K = \phi$ . [The proof is via “angles in alternate segment”.] This fact will be used for $\phi = 45^{\circ}$; (using the same diagram).

That fact will be used again for $\phi = 90^{\circ}$. (See the diagram below.)

Incorporating everything, we finally have the following picture:-

From which, we find that $DR // EA$ because $\lambda = \mu = 45^0$

$\measuredangle AXR = 90^{\circ}$ because $DR // EA$ and $\measuredangle EAX = 90^{\circ}$.

Answer 3

This is a solution of one of my friends.

File GeoGebra of this image.

We will have a coloful diagram :D

OK, at first, let $CP$ cut circle $(B)$ at $P'$, $BQ$ cut $(C)$ at $Q'$ ($P′\not\equiv P;Q′\not\equiv Q$);

$(B)$ cut $\left(C\right)$ at the 2nd point $D\implies H$ is midpoint of $AD$

We obtain $(CMPP')=(BMQQ')=-1$ (The sign of Harmonic Range)

$\implies BC, PQ, P'Q'$ meet in the same point $J$

The line through $B$ which is parallel to $AD$ cut $Q'D$ at $Q_1$, do analog thing we determine the point $P_1$.

According to Menelaus theorem we have: $\displaystyle\frac{JB}{JC}=\frac{P'M}{P'C}.\frac{Q'B}{Q'M}=\frac{DM}{CP_{1}}.\frac{BQ_{1}}{DM}=\frac{BQ_{1}}{CP_{1}}$

We will prove that $J$ be ray center of $(B)$ and $\left(C\right)\quad (*)$

Indeed: \begin{align*} (*)\displaystyle & \iff \frac{BQ_{1}}{CP_{1}}=\frac{BD}{CD}\\ &\iff \frac{\sin \angle{BDQ_{1}}}{\sin \angle{Q_{1}}}=\frac{\sin \angle{CDP_{1}}}{\sin \angle{P_{1}}}\\ &\iff \frac{\sin \angle{DAQ'}}{\sin \angle{ADQ'}}=\frac{\sin \angle{DAP'}}{\sin \angle{ADP'}}\\ &\iff \frac{DQ'}{AQ'}=\frac{DP'}{AP'}\left (=\sqrt{\frac{MD}{MA}} \right ) \text{( True )} \end{align*}

Hence we obtain $(*)$ true.

We obtain : $\overline{MP}.\overline{MP'}=\overline{MQ}.\overline{MQ'}\implies P, Q,P',Q'$ will be at the same circle.

Hence, we implies $\overline{JA}^{2}=\overline{JP'}.\overline{JQ'}=\overline{JP}.\overline{JQ}=\overline{JH}.\overline{JN}$

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Now, we use the equivalent change (This will use the result of IMO 2012)

We have : \begin{align*} &\angle AQP+\angle NAP=90^{\circ}\\ \iff &90^{\circ}-\angle AQP=\angle NAP=\angle QAP-\angle QAN\\ \iff &180^{\circ}-2\angle AQP=2\angle QAP-2\angle QAN\\ \iff &(180^{\circ}-\angle QAP-\angle AQP)-\angle AQP=\angle QAP-2\angle QAN\\ \iff &\angle APQ-\angle AQP=\angle NAP-\angle NAQ\color{red}{\quad(2)} \end{align*}

Let $E$ be intersection point of $BP$ and $CQ$. We know that $EP+EQ$ (IMO 2012), $\implies \angle EQP=\angle EPQ$.

Hence \begin{align*} (2)&\iff\angle APQ-\angle NAP=\angle AQP-\angle NAQ\\ &\iff\angle APQ+ \angle EPQ -\angle NAP=\angle AQP+\angle EQP-\angle NAQ\\ &\iff\angle CAN=\angle BAN\quad\color{red}{(3)} \end{align*}

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Solution 1

We obtain $JA\perp AN$

But from $(*)$ we obtain $AJ$ be the external bisector of $\angle A$ of $\triangle ABC$

Hence, $AN$ be the internal bisector of of $\angle A$ of $\triangle ABC$

$\implies $ OK.

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Solution 2

We easily have $JA\perp AN$, and $\triangle JAQ$ is uniforms with $\triangle JPA$ \begin{align*} &\implies \angle APQ=\angle JAQ\\ &\iff \angle APQ +\angle NAQ =90^{\circ }\\ &\iff\angle AQP +\angle NAP =90^{\circ } \end{align*}

Q.E.D