Trapezium with two right-angles: prove < AKB = 90° and more

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Trapezium with two right-angles: prove < AKB = 90° and more

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Source: Mexico 2002

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#1 h May 8, 2004, 2:12 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$ . Denote by $M$ the midpoint of the side $AB$ , and assume that $\measuredangle CMD=90^{\circ}$ . Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$ . The line $AK$ meets $BD$ at $P$ , and the line $BK$ meets $AC$ at $Q$ . Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$ .

[Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .]

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#2 h Nov 13, 2004, 8:49 PM • 2 Y

Y by Adventure10, Mango247

Proving < AKB = 90° is very easy: Since < MAD = 90° and < MKD = 90°, the points A and K lie on the circle with diameter MD, and thus < AKM = < ADM. In the right-angled triangle DAM, we have < ADM = 90° - < DMA. Thus, < AKM = 90° - < DMA. Similarly, < BKM = 90° - < CMB. Hence,

< AKB = < AKM + < BKM = (90° - < DMA) + (90° - < CMB)
= 180° - < DMA - < CMB = < CMD = 90°.

Now we are going to show $\frac{KP}{PA}+\frac{KQ}{QB}=1$ . In fact, let U be the orthogonal projection of the point K on the line AB, and let the line AC meet the line KU at T. Also let the line AK meet the line BC at Q.

Since < AKB = 90°, the point K lies on the circle with diameter AB. The center of this circle is the midpoint M of the segment AB. Hence, MK = MA = MB. Now, since MK = MB, the triangle KMB is isosceles, so that < MKB = < MBK. Thus, < CKB = 90° - < MKB = 90° - < MBK = < CBK, and it follows that the triangle KCB is isosceles, so that CK = CB. Similarly, DK = DA.

The lines AD, BC and KU are parallel to each other, since all of them are perpendicular to the line AB. Thus, from AD || BC, we have by Thales CQ : DA = CK : DK. But DK = DA; thus, CQ = CK. Together with CK = CB, this yields CQ = CB, and thus the point C is the midpoint of the segment BQ.

Since KU || BC, Thales yields TK : TU = CQ : CB. Since CQ = CB, we thus have TK = TU, and hence the point T is the midpoint of the segment KU.

Now, we have defined the point T as the point of intersection of the lines AC and KU and proved that this point T is the midpoint of the segment KU. Instead, we could have defined the point T as the midpoint of the segment KU and would be able to conclude that this point T lies on the line AC. Similarly, the same point T lies on the line BD. Altogether, the point T lies on the three lines AC, BD and KU.

Now, the points P and Q join the scene. Applying the van Aubel theorem to triangle AKB, whose cevians AQ, BP and KU concur at the point T, we get $\frac{KT}{TU} = \frac{KP}{PA}+\frac{KQ}{QB}$ . Since the point T is the midpoint of the segment KU, we have $\frac{KT}{TU}=1$ ; thus, $\frac{KP}{PA}+\frac{KQ}{QB}=1$ , and we are done.

Darij

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#3 h Nov 19, 2004, 1:40 PM • 2 Y

Y by Adventure10, Mango247

Leonardo wrote:

Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$ . Denote by $M$ the midpoint of the side $AB$ , and assume that $\measuredangle CMD=90^{\circ}$ . Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$ . The line $AK$ meets $BD$ at $P$ , and the line $BK$ meets $AC$ at $Q$ . Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$ .

We can easily prove $\angle AKB = 90^{\circ}$
We call $I$ is the intersection of $AC$ and $BD$ .
=> $\frac{IB}{IA}= \frac{AB}{CD}= \frac{BK}{CK}$
=> $IK$ // $CD$ // $AB$
=> $\frac{KP}{PA}= \frac{IK}{AD}= \frac{KC}{DC}$
$\frac{KQ}{QB}= \frac{IK}{CB}= \frac{KD}{DC}$
=> The proof is completed.

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#4 h Sep 6, 2006, 9:04 PM • 1 Y

Y by Adventure10

Prove easily that $KA\perp KB\ .$ Denote $B'\in BC\cap AK$ and $A'\in AD\cap BK\ .$ Then $CB=CB'$ and $DA=DA'\ .$

Apply the Menelaus' theorem to the transversal $\overline{AQC}$ for the triangle $BB'K$ , where $CB'=CB$ :

$\frac{AK}{AB'}\cdot \frac{CB'}{CB}\cdot\frac{QB}{QK}=1$ $\Longrightarrow$ $\frac{QK}{QB}=\frac{AK}{AB'}$ $\Longrightarrow$ $\boxed{\ \frac{QK}{QB}=\frac{AA'}{AA'+BB'}\ }\ \ (1)\ .$

Apply the Menelaus' theorem to the transversal $\overline{BPD}$ for the triangle $AA'K$ , where $DA'=DA$ :

$\frac{BK}{BA'}\cdot\frac{DA'}{DA}\cdot\frac{PA}{PK}=1$ $\Longrightarrow$ $\frac{PK}{PA}=\frac{BK}{BA'}$ $\Longrightarrow$ $\boxed{\ \frac{PK}{PA}=\frac{BB'}{AA'+BB'}\ }\ \ (2)\ .$

Therefore, from the sum of the relations $(1)$ and $(2)$ we find the required relation $\boxed{\ \frac{PK}{PA}+\frac{QK}{QB}=1\ }\ \ (*)\ .$

Remark. Denote the intersections $R\in MK\cap AC$ and $S\in MK\cap BD\ .$

Apply the Menelaus' theorem to the transversal $\overline{ARQ}$ for the triangle $MBK$ : $\frac{AM}{AB}\cdot\frac{QB}{QK}\cdot\frac{RK}{RM}=1$ $\Longrightarrow$ $\boxed{\ \frac{RK}{RM}=2\cdot\frac{QK}{QB}\ }\ \ (3)\ .$

Apply the Menelaus' theorem to the transversal $\overline{BSP}$ for the triangle $MAK$ : $\frac{BM}{BA}\cdot\frac{PA}{PK}\cdot\frac{SK}{SM}=1$ $\Longrightarrow$ $\boxed{\ \frac{SK}{SM}=2\cdot\frac{PK}{PA}\ }\ \ (4)\ .$

Therefore, from the sum of the relations $(3)$ , $(4)$ and using the relation $(*)$ obtain a new and interesting relation $\boxed{\ \frac{SK}{SM}+\frac{RK}{RM}=2\ }\ \ (**)\ .$

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#5 h Dec 23, 2022, 5:06 PM

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Let $ABCD$ be a quadrilateral with $\angle DAB=\angle ABC=90^{\circ}$ . Denote by $M$ the midpoint of the side $AB$ , and assume that $\angle CMD=90^{\circ}$ . Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$ . The line $AK$ meets $BD$ at $P$ , and the line $BK$ meets $AC$ at $Q$ . Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$ .

This post has been edited 1 time. Last edited by parmenides51, Jan 8, 2023, 12:00 AM

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#6 h Jun 14, 2023, 7:15 AM

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Since $AMKD$ and $MBCK$ are cyclic $\angle AKM = \angle ADM = 90^{\circ} - \angle AMD = \angle BMC = \angle BKC = 90^{\circ} - \angle BKM \square$ By Pappus' theorem on $A-M-B$ and $D-K-C,$ $E = AK \cap MD, F = AC \cap BD, G = MC \cap BK$ are collinear. Hence, by angle chasing $DM$ and $MC$ are perpendicular bisector of $AK$ and $BK$ , respectively. Thereby $EG \parallel AB$ .
We say that the parallel to $AB$ that passes through $K$ cut to $BD$ and $AC$ at $J$ and $L$ , respectively. Therefore $\frac{KP}{PA} + \frac{KQ}{QB} = \frac{KJ}{AB} + \frac{KL}{AB} = \frac{JL}{AB} = 1$ where the last equation is given because $\frac{JL}{AB} = \frac{FL}{AF} = \frac{EK}{AE} = 1 \blacksquare$
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