ABC right angle triangle: FK=BC

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ABC right angle triangle: FK=BC

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Source: RMO Hyderabad 2016 , P5

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kk108

#1 h Oct 12, 2016, 12:55 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Let $ABC$ be a right angled triangle with $\angle B=90^{\circ}$ . Let $AD$ be the bisector of angle $A$ with $D$ on $BC$ . Let the circumcircle of triangle $ACD$ intersect $AB$ again at $E$ ; and let the circumcircle of triangle $ABD$ intersect $AC$ again at $F$ . Let $K$ be the reflection of $E$ in the line $BC$ . Prove that $FK = BC$ .

This post has been edited 1 time. Last edited by Sayan, Oct 22, 2016, 7:06 AM
Reason: improved latex

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#2 h Oct 12, 2016, 1:42 PM • 2 Y

Y by Adventure10, Mango247

Its all about proving F,D,K are collinear?

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kk108

#3 h Oct 12, 2016, 1:46 PM • 2 Y

Y by Adventure10, Mango247

hemant1729 wrote:

Its all about proving F,D,K are collinear?

I don't have any idea , actually ...

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#4 h Oct 12, 2016, 4:03 PM • 2 Y

Y by svatejas, Adventure10

kk108 wrote:

Let ABC be a right angled triangle with B=90 degrees . Let AD be the bisector of angle A with D on BC . Let the circumcircle of triangle ACD intersect AB again in E ; and let the circumcircle of triangle ABD intersect AC again in F . Let K be the reflection of E in the line BC . Prove that FK = BC .

i have a different solution....
${\angle}BAD={\angle}CAD$
consider $(ABDF)$ .
${\angle}BAD={\angle}DAF$ ...but $BD$ and $DF$ are on the same circle.
But, $BD$ and $DF$ are chords on the same circle subtending equal angles at the circumference.
Hence, $BD$ = $DF$
Since $K$ is the reflection of $E$ in $BC$ , then, $ED$ = $KD$ .
Now consider $(AEDC)$ ,
${\angle}CAD$ = ${\angle}EAD$ = ${\angle}ECD$ .
But, $ED$ and $DC$ are chords on the same circle subtending equal angles at the circumference.
Hence, $ED$ = $DC$ = $DK$ .
therefore, $KF$ = $DK$ + $DF$ = $DC$ + $BD$ = $BC$ . HENCE PROVED!!!

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kk108

#6 h Nov 21, 2016, 11:03 AM • 2 Y

Y by Adventure10, Mango247

What does it mean to say K is the reflection of E in the line BC ?

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kk108

#7 h Dec 6, 2016, 9:07 AM • 1 Y

Y by Adventure10

hemant1729 wrote:

Its all about proving F,D,K are collinear?

Yeah sort of ........

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Vrangr

#8 h Oct 5, 2017, 6:03 AM • 4 Y

Y by AlastorMoody, Adventure10, Mango247, Rounak_iitr

Here's my proof:
$\text{Let } \angle BAC = 2\theta,$ $AD \text{ is the diameter of circle } (ABDF)$ $\implies \angle F = 90^{\circ}$ $\angle DEC = \angle DAC = \theta$ $\angle CEA = \angle ADC = 90^{\circ}-\theta$ $\implies \angle AED = 90^{\circ} + 2\theta$ $\implies \angle DEB = 90^{\circ}-2\theta$ $\implies \angle BDK = 2\theta$ $\implies \angle BDK+ \angle BDA + \angle ADF = 2\theta +2(90^{\circ}-\theta) =180^{\circ}$ $\therefore F, D, K \text{ are collinear. }$
$\because \triangle ADF \cong \triangle ADB, AF=AB$ $\because \angle F=90^{\circ} = \angle B \text{ and } \angle K = 90^{\circ} -2\theta = \angle C$ $\implies \triangle AFK \cong \triangle ABC$ $\implies BC=FK \quad \blacksquare$

This post has been edited 3 times. Last edited by Vrangr, Oct 5, 2017, 6:04 AM

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#10 h Oct 7, 2017, 11:32 AM • 2 Y

Y by Adventure10, Mango247

........................

This post has been edited 1 time. Last edited by mathlete18, Sep 12, 2019, 3:43 PM
Reason: ...

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#11 h Sep 21, 2018, 9:46 AM • 3 Y

Y by Red-Tomato, Adventure10, Mango247

mathlete18 wrote:

rd1452002 wrote:

Here's my proof:
$\text{Let } \angle BAC = 2\theta,$ $AD \text{ is the diameter of circle } (ABDF)$ $\implies \angle F = 90^{\circ}$ $\angle DEC = \angle DAC = \theta$ $\angle CEA = \angle ADC = 90^{\circ}-\theta$ $\implies \angle AED = 90^{\circ} + 2\theta$ $\implies \angle DEB = 90^{\circ}-2\theta$ $\implies \angle BDK = 2\theta$ $\implies \angle BDK+ \angle BDA + \angle ADF = 2\theta +2(90^{\circ}-\theta) =180^{\circ}$ $\therefore F, D, K \text{ are collinear. }$
$\because \triangle ADF \cong \triangle ADB, AF=AB$ $\because \angle F=90^{\circ} = \angle B \text{ and } \angle K = 90^{\circ} -2\theta = \angle C$ $\implies \triangle AFK \cong \triangle ABC$ $\implies BC=FK \quad \blacksquare$

How is AD the diameter?

Because $\Delta ABD$ is Right angled

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#12 h Sep 5, 2019, 4:33 PM • 5 Y

Y by Red-Tomato, guptaamitu1, Adventure10, Mango247, ALpHaX265

Simple angle chasing shows that $K, D, F$ are collinear with $\angle AFK=90^\circ$ . Thus, $D$ is the orthocenter of $\triangle AKC$ . Thus, $AD\perp KC$ , but $AD$ is angle bisector of $\angle BAC\implies \triangle AKC$ is isosceles. Hence, $AK=AC$ and $AK\times CB= KF\times AC=2[AKC]\implies \boxed{BC=KF}$ .

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#13 h Mar 5, 2020, 7:24 PM • 1 Y

Y by Mango247

Claim

$K,D,F$ are collinear.

Note that ,

the quadrilateral $AEDC$ is cyclic .so $\angle BED =\angle ACB =90-A$ .
Since $K$ is reflection of $E$ in $BC$ hence $EB=BK$ . Since $DB \perp EK$ we have $\angle EDB =\angle BDK=A$ .

So, $\angle EDK =2A$ .
Again $AEDF$ is concyclic . So $\angle EDF=180-\angle BFC=180-2A$ .

So , $\angle KDF =180$ .

Our claim is proved $\blacksquare$ .

$AB =AD \sin \frac{A}{2} = AF$ .
Note that $\triangle ABC \sim \triangle AKF$ .

$\implies \frac{AB}{AF} =\frac{AK}{AC}$ .

$\implies \boxed{AK=AC}$ .

Finally we are getting $\triangle ABC \cong \triangle AKF$ .

$\implies \boxed{BC=KF}$ .

This post has been edited 3 times. Last edited by ftheftics, Mar 5, 2020, 7:25 PM
Reason: Nn

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#14 h Oct 15, 2020, 5:31 PM

Y by

polarLines wrote:

kk108 wrote:

Let ABC be a right angled triangle with B=90 degrees . Let AD be the bisector of angle A with D on BC . Let the circumcircle of triangle ACD intersect AB again in E ; and let the circumcircle of triangle ABD intersect AC again in F . Let K be the reflection of E in the line BC . Prove that FK = BC .

i have a different solution....
${\angle}BAD={\angle}CAD$
consider $(ABDF)$ .
${\angle}BAD={\angle}DAF$ ...but $BD$ and $DF$ are on the same circle.
But, $BD$ and $DF$ are chords on the same circle subtending equal angles at the circumference.
Hence, $BD$ = $DF$
Since $K$ is the reflection of $E$ in $BC$ , then, $ED$ = $KD$ .
Now consider $(AEDC)$ ,
${\angle}CAD$ = ${\angle}EAD$ = ${\angle}ECD$ .
But, $ED$ and $DC$ are chords on the same circle subtending equal angles at the circumference.
Hence, $ED$ = $DC$ = $DK$ .
therefore, $KF$ = $DK$ + $DF$ = $DC$ + $BD$ = $BC$ . HENCE PROVED!!!

This solution is incomplete as you first need to prove that points K,D,F are collinear in order to write KF = DK + DF

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#15 h Oct 15, 2020, 5:43 PM

Y by

ftheftics wrote:

Claim

$K,D,F$ are collinear.

Note that ,

the quadrilateral $AEDC$ is cyclic .so $\angle BED =\angle ACB =90-A$ .
Since $K$ is reflection of $E$ in $BC$ hence $EB=BK$ . Since $DB \perp EK$ we have $\angle EDB =\angle BDK=A$ .

So, $\angle EDK =2A$ .
Again $AEDF$ is concyclic . So $\angle EDF=180-\angle BFC=180-2A$ .

So , $\angle KDF =180$ .

Our claim is proved $\blacksquare$ .

$AB =AD \sin \frac{A}{2} = AF$ .
Note that $\triangle ABC \sim \triangle AKF$ .

$\implies \frac{AB}{AF} =\frac{AK}{AC}$ .

$\implies \boxed{AK=AC}$ .

Finally we are getting $\triangle ABC \cong \triangle AKF$ .

$\implies \boxed{BC=KF}$ .

Triangle ABC is not congruent to triangle AKF,
but instead the it should have been
triangle ABC is congruent to triangle AFK

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#16 h Oct 15, 2020, 5:46 PM • 1 Y

Y by Mango247

Vrangr wrote:

Here's my proof:
$\text{Let } \angle BAC = 2\theta,$ $AD \text{ is the diameter of circle } (ABDF)$ $\implies \angle F = 90^{\circ}$ $\angle DEC = \angle DAC = \theta$ $\angle CEA = \angle ADC = 90^{\circ}-\theta$ $\implies \angle AED = 90^{\circ} + 2\theta$ $\implies \angle DEB = 90^{\circ}-2\theta$ $\implies \angle BDK = 2\theta$ $\implies \angle BDK+ \angle BDA + \angle ADF = 2\theta +2(90^{\circ}-\theta) =180^{\circ}$ $\therefore F, D, K \text{ are collinear. }$
$\because \triangle ADF \cong \triangle ADB, AF=AB$ $\because \angle F=90^{\circ} = \angle B \text{ and } \angle K = 90^{\circ} -2\theta = \angle C$ $\implies \triangle AFK \cong \triangle ABC$ $\implies BC=FK \quad \blacksquare$

There is a small silly error in this solution, namely the line
"∠CEA = ∠ADC = 90 - theta" is not correct and instead it should have been
"∠CEA = ∠ADC = 90 + theta"

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