INMO 2019 P1

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474 posts

#1 h Jan 20, 2019, 12:15 PM • 12 Y

Y by AlastorMoody, integrated_JRC, saksham1729, nguyendangkhoa17112003, RudraRockstar, Purple_Planet, mathematicsy, michaelwenquan, ImSh95, Adventure10, Mango247, Rounak_iitr

Let $ABC$ be a triangle with $\angle{BAC} > 90$ . Let $D$ be a point on the segment $BC$ and $E$ be a point on line $AD$ such that $AB$ is tangent to the circumcircle of triangle $ACD$ at $A$ and $BE$ is perpendicular to $AD$ . Given that $CA=CD$ and $AE=CE$ . Determine $\angle{BCA}$ in degrees.

This post has been edited 2 times. Last edited by div5252, Jan 20, 2019, 5:54 PM

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Lightprince

1 post

Lightprince

#2 h Jan 20, 2019, 12:21 PM • 4 Y

Y by Subh108, michaelwenquan, ImSh95, Adventure10

i got 60 as answer

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AlastorMoody

2125 posts

AlastorMoody

#3 h Jan 20, 2019, 12:25 PM • 10 Y

Y by ILIILIIILIIIIL, alex_g, MathPassionForever, BobaFett101, ayan_mathematics_king, Arhaan, Tris, Purple_Planet, ImSh95, Adventure10

My solution (motivated from Tumon2001's hint!):
Let $\angle BCA=x$ and Let $AE \cap \odot (ABC) =X$ , then, $\angle BXA=\angle BCA=x=\angle BAX$ $\implies AB=BX \text{ and } AE=EX=EC \implies (E) \equiv \odot (ABC) \implies \angle ABX=180^{\circ}-2x=90^{\circ} \implies \boxed{x= 45^{\circ}}$

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MathPassionForever

1663 posts

MathPassionForever

#4 h Jan 20, 2019, 1:36 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

AlastorMoody wrote:

My solution (motivated from Tumon2001's hint!):
Let $\angle BCA=x$ and Let $AE \cap \odot (ABC) =X$ , then, $\angle BXA=\angle BCA=x=\angle BAX$ $\implies AB=BX \text{ and } AE=EX=EC \implies (E) \equiv \odot (ABC) \implies \angle ABX=180^{\circ}-2x=90^{\circ} \implies \boxed{x= 45^{\circ}}$

AE=EX?

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AlastorMoody

2125 posts

AlastorMoody

#5 h Jan 20, 2019, 1:46 PM • 4 Y

Y by Purple_Planet, ImSh95, Adventure10, Mango247

@above, since, $BE \perp AX$ and we already have $AB=BX \implies \Delta ABX$ is isosceles, hence, $AE=EX$

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TheDarkPrince

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TheDarkPrince

#6 h Jan 20, 2019, 1:53 PM • 6 Y

Y by alex_g, Purple_Planet, BVKRB-, ImSh95, Adventure10, Mango247

Bash bash bash

Simple trig

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MathPassionForever

1663 posts

MathPassionForever

#7 h Jan 20, 2019, 1:53 PM • 4 Y

Y by Aryan-23, ImSh95, Adventure10, Mango247

AlastorMoody wrote:

@above, since, $BE \perp AX$ and we already have $AB=BX \implies \Delta ABX$ is isosceles, hence, $AE=EX$

Ok, missed that.
One doubt - I computed all the angles in the figure without construction, but could not reach the answer.
How many would I get for this?

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JRyno

361 posts

JRyno

#8 h Jan 20, 2019, 2:01 PM • 2 Y

Y by ImSh95, Adventure10

If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get?

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aadhitya

98 posts

aadhitya

#9 h Jan 20, 2019, 2:08 PM • 6 Y

Y by Tris, Purple_Planet, ImSh95, Adventure10, Mango247, TensorGuy666

Here's my solution(SYNTHETIC

):
Let $\angle ADC = x$ .
After some angle chase we get,
$\angle ABC = \angle BCE = 3x - 180.$ . So, $AB \parallel CE$ .
Now extend $BA$ to $X$ such that $XC \parallel AE$ .
Notice that $AECX$ is a rhombus. Therefore $\angle EXC = 90 - x = \angle BEC$ . Therefore, $BEXC$ is cyclic.
$\implies \angle XBC = \angle XEC.$
Therefore, $3x - 180 = 90 - x$ .
$\implies \angle ACB = 180 - 2x = 45$

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MathPassionForever

1663 posts

MathPassionForever

#10 h Jan 20, 2019, 2:09 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

JRyno wrote:

If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get?

Dunno, maybe 14-15 in the worst case?

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Kayak

1298 posts

Kayak

#11 h Jan 20, 2019, 2:12 PM • 8 Y

Y by AlastorMoody, Math-Ninja, biomathematics, SHREYAS333, amar_04, ImSh95, Adventure10, Mango247

MathPassionForever wrote:

JRyno wrote:

If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get?

Dunno, maybe 14-15 in the worst case?

Wtf

2-3 in the best case

I got a biquadratic equation in cos phi before 40 seconds the exam ended and then rage quotes.

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MathPassionForever

1663 posts

MathPassionForever

#12 h Jan 20, 2019, 2:15 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Kayak wrote:

MathPassionForever wrote:

JRyno wrote:

If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get?

Dunno, maybe 14-15 in the worst case?

Wtf

2-3 in the best case

I got a biquadratic equation in cos phi before 40 seconds the exam ended and then rage quotes.

My friend PMed me the one he got, it was $cos(2x)\cdot cos(x) = -cos(2x)$

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Tumon2001

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Tumon2001

#13 h Jan 20, 2019, 2:39 PM • 7 Y

Y by AlastorMoody, RudraRockstar, Purple_Planet, GeoKing, ImSh95, Adventure10, Mango247

Looks like someone has already posted my solution above (post #3)

.....to reduce it even further...

Let $AE\cap \odot (ABC)=F$ . Angle chasing gives $BA=BF\implies EA=EF$ . Since, $EA=EC$ , so $E$ is the center of $\odot (ABC)$ . Thus, $\angle ACB=\frac {1}{2}\angle AEB=45^{\circ}$ .

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Yagami1728

308 posts

Yagami1728

#14 h Jan 20, 2019, 4:28 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Kayak wrote:

MathPassionForever wrote:

JRyno wrote:

If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get?

Dunno, maybe 14-15 in the worst case?

Wtf

2-3 in the best case

I got a biquadratic equation in cos phi before 40 seconds the exam ended and then rage quotes.

Really?
Surely you would get more? Assuming it can be simplified in just 1-2 more steps...

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Arhaan

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Arhaan

#15 h Jan 20, 2019, 4:41 PM • 2 Y

Y by ImSh95, Adventure10

^ Evaluators are not so compassionate in giving partials. As TDP said the point slabs appear to be 0,1,12,17.

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gulkr10sept

27 posts

gulkr10sept

#52 h Jan 31, 2021, 3:19 PM • 1 Y

Y by ImSh95

Lightprince wrote:

i got 60 as answer

Then you are wrong because then..... angle B would be 0 degree.....

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Maths_1729

390 posts

Maths_1729

#53 h Feb 9, 2021, 8:49 AM • 1 Y

Y by ImSh95

Another Solution can be..
First observe By Tangent-Secant theorem we have $\angle BAE=\angle BCA$ Now we will choose a variable Point $B^*\in BE$ Such $EB^*=AD=AC$ Now there are $2$ cases
Case 1-: if $B^*E<BE$
Then observe $\angle B^*CA=\angle B^*AE=45^\circ \implies$ $\angle BCA+\angle BCB^*=\angle BAE -\angle BAB^*$ Which is only possible if $\angle BCB^*=\angle BAB^*=0^\circ$ or $B^*\equiv B$ Hence $\angle BCA=45^\circ$ in this case.
Case 2-: if $B^*E>BE$
Then $\angle B^*AE=\angle BAE +\angle B^*AB=\angle BCA-\angle B^*CB$ And here also it is only possible if $\angle B^*AB=\angle B^*CB=0^\circ$
Hence $B^*\equiv B$ and $\angle BCA=45^\circ$ $\blacksquare$

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MrOreoJuice

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MrOreoJuice

#54 h Apr 30, 2021, 4:16 PM • 2 Y

Y by PRMOisTheHardestExam, ImSh95

Simple solution:
Let $\angle ADC = 2x$ and $F$ be the foot of perpendicular from $A$ to $BC$ and $G = EX \cap AC$ .
$\implies ABEF$ is cyclic.
Simple angle chase done in picture

$\angle GFA = \angle GAF \implies AG=GF$ and $GF = GC \implies G$ is midpoint of $AC \implies \angle EGA = 90^\circ$
$\implies 2x + 2x = 90^\circ \implies 2x = 45^\circ = \angle BCA$

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nabodorbuco

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nabodorbuco

#55 h Aug 19, 2021, 5:25 PM • 1 Y

Y by ImSh95

Hi there! I just made a video about it. Hope you like it!

https://youtu.be/UVFx1gVWP4w

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Sprites

478 posts

Sprites

#56 h Aug 19, 2021, 5:42 PM • 1 Y

Y by ImSh95

nabodorbuco wrote:

Hi there! I just made a video about it. Hope you like it!

https://youtu.be/UVFx1gVWP4w

Very well explained!!
Solution

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Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#57 h Mar 5, 2022, 1:25 PM • 1 Y

Y by ImSh95

we have $\angle ECA = \angle EAC = \angle ADC = x$ . $\angle BAE = \angle 180 - 2x = \angle AEC$ so $CE || AB$ . Let line from $C$ parallel with $AE$ meet $AB$ at $S$ . we have $\angle CSE = \angle 90 - x = \angle CBE$ so $BECS$ is cyclic so $\angle 180 = \angle EBS + \angle ECS = 2x - 90 + 2x$ so $4x = 270$ so $x = 67.5$ . $\angle BCA = \angle DCA = \angle 180 - 2x = 45$ .
we're Done.

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Taco12

1757 posts

Taco12

#58 h Mar 8, 2023, 1:06 AM • 1 Y

Y by ImSh95

Let $\angle BCA = 2x$ . Notice that by angle chasing, we can directly get the measure of all angles in the problem statement:
$\begin{align*} \angle CAE = \angle ADC = \angle BDE &= 90-x \\ \angle BCA = \angle BAD &= 2x \\ \angle ABD = \angle BCE &= 90-3x \\ \angle CBE &= x \\ \angle ADB = \angle CDE &= 90+x. \\ \end{align*}$
Now, by Quadrilateral Ratio Lemma, we have $\frac{\sin\angle BAE}{\sin\angle CAE}=\frac{\sin\angle ABE}{\sin\angle ACE}\cdot\frac{\sin\angle BCE}{\sin\angle CBE} \leftrightarrow \sin(2x)\sin(x)=\sin(90-2x)\sin(90-3x)\leftrightarrow x=22.5^{\circ} \rightarrow 2x=\boxed{45^{\circ}}.$

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lifeismathematics

1188 posts

lifeismathematics

#59 h Jun 30, 2023, 1:18 PM • 2 Y

Y by ImSh95, Rounak_iitr

We extend $BA$ to point $P$ such that $CP \parallel AE$

this gives $AECP$ to be a rhombus

denote $\angle{ADC}=\theta$ a, since $BE \perp AD$ we have $\angle{DBE}=90^{\circ}-\theta$ , also we have $\angle{ADC}=\angle{DAC}=\angle{ECA}=\theta$ we have $\angle{DCA}=180^{\circ}-2\theta \implies \angle{ECD}=3\theta-180^{\circ}$

since $AECP$ is a rhombus we have $\angle{EPC}=90^{\circ}-\theta \implies BECP$ is a cyclic quadrilateral since we have $\angle{DBE}=90^{\circ}-\theta$

so we have $\angle{ECD}=\angle{APE} \implies 3\theta-180^{\circ}=90^{\circ}-\theta \implies 180-2\theta=45^{\circ}$

so $\angle{BCA}=180^{\circ}-2\theta=\boxed{45^{\circ}}$ $\blacksquare$

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Spectator

657 posts

Spectator

#60 h Jul 20, 2023, 1:04 AM

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Bruh

Let $\alpha = \angle{DAC}$ . It suffices to find $180-2\alpha$ . Note that $\angle{AEC} = 180-2\alpha = \angle{BAE}$ , because of tangent properties. Define $X$ as the foot of the altitude from $C$ onto $AD$ . Note that $\triangle{ABE}\sim \triangle{ECX}$ and also $\triangle{CXD} \sim \triangle{BED}$ . Because of this, we have
$\frac{XD}{DE} = \frac{XE}{AE}\implies \frac{AD}{2\cdot DE} = \frac{AD+2\cdot DE}{2\cdot (AD+DE)} \implies AD = \sqrt{2}\cdot DE$ and we also have
$\frac{AB}{AE}=\frac{AB}{CE} = \frac{CX}{BE} = \frac{XD}{DE} \implies AB = \sqrt{2}AE$ Thus, $180-2\alpha = \boxed{45^{\circ}}$ .

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cj13609517288

1891 posts

cj13609517288

#61 h Oct 31, 2023, 2:16 AM

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Let $F$ be the other intersection of $(ACD)$ and $CE$ and let $\alpha=\angle ADC$ . Then
$\alpha=\angle ADC=\angle DAC=\angle ACE,$ so
$\angle BAC+\angle ACE=\angle BAD+2\alpha=\angle ACD+\angle DAC+\angle ADC=180^{\circ},$ so $AB\parallel CE$ and $AC\parallel DF$ .
Let $P$ be the foot of the perpendicular from $C$ to $AD$ . Note that $AP=PD$ . We have $\triangle BED\sim\triangle CPD$ by AA similarity.

Now consider the negative homothety $h$ centered at $D$ such that $h(B)=C$ . Then by the similarity, $h(E)=P$ . We have $h(AB)=CE$ since they are parallel. Since $h(AE)=AE$ , $h(A)=h(AB\cap AE)=CE\cap AE=E$ . So $DE^2=(DA)(DP)=\frac12 DA^2$ , so $\sqrt2 DE=DA$ .

Now let $DF=a$ . Then $AC=(\sqrt2+1)a$ , so $CD=AF=(\sqrt2+1)a$ . If we let $AD=b$ , applying Ptolemy's on cyclic quadrilateral $ADFC$ gives
$(\sqrt2+1)a^2+b^2=(\sqrt2+1)^2 a^2\Longrightarrow b=\sqrt{2+\sqrt2} a\Longrightarrow DE=\sqrt{4+2\sqrt2}\cdot\frac{a}{2}.$ Thus
$\cos\alpha=\cos\angle FDE=\frac{a/2}{DE}=\frac{1}{\sqrt{4+2\sqrt2}}\Longrightarrow \cos(2\alpha)=\frac{2}{4+2\sqrt2}-1=\frac{1}{2+\sqrt2}-1=\frac{2-\sqrt2}{2}-1=\frac{-\sqrt2}{2}$ $\Longrightarrow \cos(\angle ACB)=\frac{\sqrt2}{2}\Longrightarrow \boxed{\angle ACB=45^{\circ}}.$

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SilverBlaze_SY

66 posts

SilverBlaze_SY

#62 h Jan 10, 2024, 8:26 AM

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Let $AX\perp BC$ such that $X$ lies on $BC$ . Extending $EX$ to meet $AC$ at $F$ .
Let $\angle BCA= 2\theta \Rightarrow \angle CAD= \angle CDA= 90^{\circ}-\theta$ . (As $\Delta CAD$ is isosceles.)
Also, $AE=CE \Rightarrow \angle EAC= \angle ACE = 90^{\circ}-\theta \Rightarrow \angle AEC= 2\theta$ .
As $AB$ is tangent to the circumcircle of $\Delta CAD \Rightarrow \angle DAB = \angle ACD= 2\theta$ . (by alternate segment theorem).
From the previous results, $\angle AEC=\angle EAB = 2\theta \Rightarrow AB \parallel CE$ .

$\angle BEA = \angle BXA = 90^{\circ} \Rightarrow ABEX$ is cyclic.
$\Rightarrow \angle BAE= \angle BXE= 2\theta = \angle CXF$ (vertically opposite).
Thus, $\Delta CFX$ is isosceles $\Rightarrow XF=CF$ .

$\angle AXF= 90^{\circ} -\angle CXF= 90^{\circ}-2\theta = \angle XAF$ .
Thus, $\Delta AXF$ is isosceles $\Rightarrow AF=XF$ .
But $XF=CF\Rightarrow AF=CF$ . Also, $AE=CE \Rightarrow \Delta AEF \cong \Delta CEF$ .
$\Rightarrow \angle AFE= 90^{\circ} \Rightarrow \angle FAX+ \angle FXA= 90^{\circ} \Rightarrow 2(90^{\circ}-2\theta)=90^{\circ}$
$\Rightarrow 2\theta=45^{\circ} \Rightarrow \angle BCA= \boxed{45^{\circ}}$

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peppapig_

280 posts

peppapig_

#63 h Mar 14, 2024, 9:06 PM

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Let $\angle ACD=2a$ .

C1: I claim that $AB\parallel CE$ . This is because since $AB$ is tangent to $(ADC)$ at $A$ , we have that
$\angle BAE=\angle ACD=2a.$ Additionally, since $AE=CE$ , we have that
$\angle AEC=180-2\angle EAC=180-2\angle DAC=\angle ACD=2a,$ since $CA=CD$ . Therefore, since $\angle BAE=\angle AEC$ , we have that $AB\parallel CE$ , proving our claim.

C2: I claim that $\triangle BDA\sim \triangle CDE$ . This follows directly from (C1). Since $AB\parallel CE$ , this gives us that
$\angle ABD=\angle ECD,$ and
$\angle BAD=\angle DEC,$ meaning that $\triangle BDA\sim \triangle CDE$ , proving our claim.

C3: Let $X\neq D$ be on line $AB$ so that $EX=DX$ . I claim that $\triangle BDX\sim CDA$ . Note that since $\angle BED=90$ and $ED=DX$ , we have that $\triangle BDX$ is isosceles with $BD=BX$ . Since $\triangle CDA$ is also isosceles with $CA=CD$ , to prove that they're similar, we just need to prove that $\angle BDE=\angle CDA$ . This is obvious since lines $AE$ and $BC$ intersect at $D$ . Therefore $\triangle BDX\sim CDA$ , proving our claim.

C4: I claim that $\frac{AD}{DE}=\sqrt{2}$ . This is because by (C2), we have that $\triangle BDA\sim \triangle CDE$ , meaning that
$\frac{BD}{DC}=\frac{AD}{DE}.$ Additionally, by (C3), we also have that $\triangle BDA\sim \triangle CDE$ , meaning that
$\frac{BD}{DC}=\frac{XD}{AD}=\frac{2*DE}{AD}.$ Equating the two equations gives that
$2*\frac{DE}{AD}=\frac{AD}{DE},$ implying that $\frac{AD}{DE}=\sqrt{2}$ , proving our claim.
C5: I claim that $\frac{EA}{AB}=\frac{\sqrt{2}}{2}$ . This is because
$\frac{EA}{AB}=\frac{EC}{AB}=\frac{ED}{AD}=\frac{\sqrt{2}}{2},$ by (C4), proving our claim.

Finally, notice that $\frac{\sqrt2}{2}\frac{EA}{AB}=\cos(2a)$ , meaning that
$\angle BCA=\angle ACD=2a=45,$ finishing the problem.

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Mathandski

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Mathandski

#64 h Aug 23, 2024, 11:15 PM

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Subjective Rating (MOHs) $\text{[math]}$

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TestX01

339 posts

TestX01

#65 h Aug 24, 2024, 2:00 AM

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Sketch:

Angle chase for $AC$ tangent to $(CDE)$ . Invert at $A$ preserving $(CDE)$ . Then we see that $D\leftrightarrow E$ , $C$ is fixed, and that $B$ is sent to the foot of $D$ onto $AB$ . Note that $ACEB'$ is cyclic. We reverse reconstruct $E$ . Let $E$ be $AD\cap (ACB')$ . Then $(CDE)$ is tangent to $AB$ . Note that by angle chase,
$\measuredangle DAB'=\measuredangle BCA=\measuredangle AEC=\measuredangle AB'C$ Letting $CB'\cap AD=X$ , we see by Thales that $X$ is the midpoint of $AD$ . However, $CX$ is then a symmedian of $\triangle CAD$ (Isos triangle). This means that $B'D$ is tangent to $(ACD)$ . Clearly, we can get $\angle BCA=45^\circ$ by alternate segment theorem.

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L13832

262 posts

L13832

#67 h Nov 21, 2024, 11:51 AM • 1 Y

Y by CRT_07

Solution

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