6 points in a parallelogram, two equal segments

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6 points in a parallelogram, two equal segments

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#1 h Oct 13, 2017, 6:53 PM • 4 Y

Y by MathAllTheWay, nguyendangkhoa17112003, Adventure10, Mango247

Let $ABCD$ be a parallelogram. $P \in (CD), Q \in (AB)$ , $M= AP \cap DQ$ , $N=BP \cap CQ$ , $K=MN \cap AD$ , $L= MN \cap BC$ . Prove that $BL=DK$ .

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#2 h Jan 8, 2019, 1:25 PM • 1 Y

Y by Adventure10

By Pappus Theorem we have that $M,N$ and $S=AC\cap BD$ are collinear. From here it is obvious $BL=DK$ since $\bigtriangleup DKS \cong \bigtriangleup SLB$ (by $ASA$ )

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#3 h Jan 8, 2019, 3:18 PM • 3 Y

Y by RAMUGAUSS, Mathlover_1, Adventure10

Extend $KL$ and let it meet $AB$ at $X$ .
Menalaus theorem on triangles $ADQ$ , $APQ$ , $QCB$ gives
$\frac {DM}{MQ} \frac{QX}{AX} \frac {AK}{KD}=1$ .
$\frac {DN}{NB} \frac {BX}{AX} \frac {AM}{MP}=1$ $\frac {CL}{LB} \frac {BX}{QX} \frac {QN}{NC}=1$ and for parallel sides $\frac {AM}{MQ}=\frac {DM}{MQ}$ and $\frac {QN}{NC}=\frac {NB}{PN}$ solving and some coputation, I think we will get the result

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#4 h Dec 28, 2023, 4:23 AM

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Let $AC \cap BD = O$ , clearly $BO = DO$ . If it turns out that $O$ lies on the line $MN$ , then it would follow that $\triangle DKO \cong \triangle BLO$ by side and two angles and hence $BL = DK$ .

Choose coordinates so that $A(0,0)$ , $B(1,0)$ and $C(a,b)$ - then $D(a-1,b)$ , as the segments $AB$ and $CD$ are equal and parallel. Let $Q(q,0)$ and $P(p,b)$ . We may compute $M$ and $N$ via the equations of the lines $AP$ , $DQ$ , $BP$ , $CQ$ , but let us be smarter. By Thales' theorem we have $\frac{AM}{MP} = \frac{AQ}{DP} = \frac{q}{p-a+1}$ , so $M\left(\frac{0+\frac{q}{p-a+1}p}{1+\frac{q}{p-a+1}},\frac{0+\frac{q}{p-a+1}b}{1+\frac{q}{p-a+1}}\right) = M\left(\frac{pq}{p+q-a+1},\frac{bq}{p+q-a+1}\right).$ Analogously $\frac{QN}{NC} = \frac{BQ}{CP} = \frac{1-q}{a-p}$ , whence
$N\left(\frac{q+\frac{1-q}{a-p}a}{1 + \frac{1-q}{a-p}}, \frac{0 + \frac{1-q}{a-p}b}{1+\frac{1-q}{a-p}}\right) = N\left(\frac{a-pq}{a-p-q+1},\frac{(1-q)b}{a-p-q+1}\right).$ Hence as $O\left(\frac{a}{2},\frac{b}{2}\right)$ , the slopes of the lines $OM$ and $ON$ are
$\frac{\frac{bq}{p+q-a+1} - \frac{b}{2}}{\frac{pq}{p+q-a+1} - \frac{a}{2}} = \frac{b (a - p + q - 1)}{a^2 - a(p + q + 1) + 2 p q}, \ \ \frac{\frac{(1-q)b}{a-p-q+1} - \frac{b}{2}}{\frac{a-pq}{a-p-q+1} - \frac{a}{2}} = \frac{b (a - p + q - 1)}{a^2 - a (p + q + 1) + 2 p q}$ i.e. they are equal and since the lines have a common point (namely $O$ ), they coincide. This completes the proof.

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