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

hemant1729 wrote: Its all about proving F,D,K are collinear? I don't have any idea , actually ...

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!!!

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

hemant1729 wrote: Its all about proving F,D,K are collinear? Yeah sort of ........

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

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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

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 $AK\times CB= KF\times AC=2[AKC]\implies \boxed{BC=KF}$.

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 $\triangle ABC \cong \triangle AKF$. $\implies \boxed{BC=KF}$.

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

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

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"