Parallelogram and similar triangles

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Parallelogram and similar triangles

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#1 h Oct 22, 2005, 4:50 PM • 2 Y

Y by Adventure10, Mango247

Isosceles, similar triangles $QPA$ and $SPB$ are constructed (outwards) on the sides of parallelogram $PQRS$ (where $PQ = AQ$ and $PS = BS$ ). Prove that triangles $RAB$ , $QPA$ and $SPB$ are similar.

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#2 h Dec 17, 2005, 3:50 PM • 2 Y

Y by Adventure10, Mango247

$\Delta QPA \sim \Delta PBS$ .
$QR = PS = BS$ . $QA = PQ = RS$ . $\measuredangle AQR = \measuredangle BSR$ . Hence $\Delta AQR \equiv \Delta BSR$ .
$180^{\circ} - (\measuredangle QAR + \measuredangle QRA) = \measuredangle AQP + \measuredangle PQR$ .
$180^{\circ} - \measuredangle QPS = \measuredangle PQR$ .
Subtracting these $\measuredangle QPS - (\measuredangle QAR + \measuredangle QRA) = \measuredangle AQP = \measuredangle QRS - (\measuredangle QRA + \measuredangle BRS) = \measuredangle ARB$ .
Also $AR = RB$ . Hence $\Delta QPA \sim \Delta PBS \sim \Delta ARB$ .

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#3 h Dec 15, 2013, 12:11 PM • 2 Y

Y by Adventure10, Mango247

Note that $AQ=PQ=RS$ so $QASR$ is isosceles trapezoid,similarly $QRSB$ also.hence pentagon $AQRSB$ is cyclic,also $AQ=RS,QR=SB$ and $\angle AQR=\angle RSB$ because $\angle AQP=\angle BSP$ ,hence $AQR$ and $RSB$ are congruent. so $RA=RB$ ,also $\angle ABQ=\angle RBS$ ( $AQ=RS$ ),so $\angle ABR=\angle PBS$ ,hence we are done.

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#4 h Dec 16, 2013, 2:02 PM • 2 Y

Y by Adventure10, Mango247

vanu1996 wrote:

Note that $AQ=PQ=RS$ so $QASR$ is isosceles trapezoid,similarly $QRSB$ also.hence pentagon $AQRSB$ is cyclic,also $AQ=RS,QR=SB$ and $\angle AQR=\angle RSB$ because $\angle AQP=\angle BSP$ ,hence $AQR$ and $RSB$ are congruent. so $RA=RB$ ,also $\angle ABQ=\angle RBS$ ( $AQ=RS$ ),so $\angle ABR=\angle PBS$ ,hence we are done.

To revive such an old thread with a wrong solution... how did you get isosceles trapezoids??

Best regards,
sunken rock

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#5 h Dec 16, 2013, 4:09 PM • 2 Y

Y by Adventure10, Mango247

Let $AR \cap PQ=L$ ,we can easily prove that $\angle QAL=\angle SRB$ and $AR=RB$ ,Now $\angle QAL+\angle AQL+\angle ALQ=180=\angle RLP+\angle LRB+\angle SRB$ hence $\angle AQP=\angle ARB$ .
@sunken,I am sorry.

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#6 h Dec 31, 2021, 8:21 AM

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Let $\angle QAP=x$ and $\angle PSB=y$ . As we have $\Delta QPA\sim \Delta SPB$ then,
$\frac{PA}{PB}=\frac{PQ}{BS} ...(i)$ We have $\Delta BSR\cong \Delta RQA$ , hence we get $RA=RB$ .

Let us assume $\angle RBS=b$ and $\angle RAQ=a$ , then, we get $\angle RSB=\angle RQA=180-(a+b)=z$ . Let the intersection of $RB$ and $PS$ be named as $K$ , then let $\angle RKS=k$ . Hence we get $\angle SPQ=k+a$ and assume

Assume,
$360-(k+a)-2x=z=\angle RSP+y$ $180=y+2x$ which is true, so we get $\angle APB=z$ and $\Delta RSB\sim \Delta APB$ , from (i), therefore we have $\angle RAB=x-a+a=x$ .

Hence,
$\Delta QPA\sim\Delta SPB\sim\Delta RAB$

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