prove that the rectangle is a square, double angle is given

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prove that the rectangle is a square, double angle is given

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#1 h Sep 8, 2018, 11:26 PM • 3 Y

Y by megarnie, Adventure10, Mango247

Let $ABCD$ be a rectangle with $AB\ge BC$ Point $M$ is located on the side $(AD)$ , and the perpendicular bisector of $[MC]$ intersects the line $BC$ at the point $N$ . Let ${Q} =MN\cup AB$ . Knowing that $\angle MQA= 2\cdot \angle BCQ$ , show that the quadrilateral $ABCD$ is a square.

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#3 h May 11, 2020, 9:25 AM

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Let point $E$ be point on side $BN$ and point $F$ on side $AM$ such that $\angle FQM=\angle FQA=\angle BQE=\angle EQN=\angle QCB$ .
Since this is a rectangle, we can obviously see that triangle $EQC$ is right angle triangle, so $\angle EQC=\angle FQC=90^\circ$
Since we have perpendicular bisector of $MC$ , triangle $NMC$ is isosceles.
Let point $G$ be on side $MQ$ such that $CG \perp MQ$ , since $CGQB$ is cyclic, $BC=CG$
Let point $H$ be on side $BC$ such that $MH \perp BC$ and $MH=AB$ , since $NMC$ is isosceles, this means that $MH=CG=BC=AB$
So we proved that $AB=BC$ meaning that this rectangle is a square.

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#4 h May 11, 2020, 10:56 AM

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Clearly $\angle CMN=\angle MCN=\angle CMB\ (\ 1\ )$ .
Let $m(\widehat{BCQ})=\alpha$ , then $\angle BQC=90^\circ-\alpha=\angle CQM\ (\ 2\ )$ , thus $C$ is $A-$ excenter of $\triangle AQM$ and $AC$ is the angle bisector of $\angle BAD$ , that is, $ABCD$ is a square.

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