Show $measuredangle ATB=measuredangle CTA$ in excircle configuration

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I recently came across the following problem in olympiad training material:

A circle $k$ is internally tangent to sides $AB, AC$ of $Delta ABC$ and its circumcircle in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of $XY$ and $AZ$. Prove that $measuredangle XTB=measuredangle CTY$.

By inverting through the circle with center $A$ that is passing through $T$, I managed to transform the problem into the following, in my opinion easier, version:

The $A$-excircle of $Delta ABC$ touches $AB, AC$ and $BC$ in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of the circumcircle of $Delta AXY$ with $AZ$. Prove $measuredangle ATB=measuredangle CTA$.

However, I have not been able to solve the second version of the problem either and only got as far as $measuredangle ATX=measuredangle YTA$ which leaves $measuredangle YTC=measuredangle BTX$ to prove.
Furthermore, the circle through $A, X $ and $Y $ is obviously Thales' circle with diameter $AM $ where $M $ is the center of the excircle.

Does anyone have a suggestion how to proceed or how to solve the initial problem without an inversion?

geometry contest-math circle triangle

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edited Jul 17 at 8:40

asked Jul 16 at 20:18

mxian

563216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you add a diagram?
  – Moti
  Jul 17 at 5:49
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

I recently came across the following problem in olympiad training material:

A circle $k$ is internally tangent to sides $AB, AC$ of $Delta ABC$ and its circumcircle in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of $XY$ and $AZ$. Prove that $measuredangle XTB=measuredangle CTY$.

By inverting through the circle with center $A$ that is passing through $T$, I managed to transform the problem into the following, in my opinion easier, version:

The $A$-excircle of $Delta ABC$ touches $AB, AC$ and $BC$ in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of the circumcircle of $Delta AXY$ with $AZ$. Prove $measuredangle ATB=measuredangle CTA$.

However, I have not been able to solve the second version of the problem either and only got as far as $measuredangle ATX=measuredangle YTA$ which leaves $measuredangle YTC=measuredangle BTX$ to prove.
Furthermore, the circle through $A, X $ and $Y $ is obviously Thales' circle with diameter $AM $ where $M $ is the center of the excircle.

Does anyone have a suggestion how to proceed or how to solve the initial problem without an inversion?

geometry contest-math circle triangle

share|cite|improve this question

edited Jul 17 at 8:40

asked Jul 16 at 20:18

mxian

563216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you add a diagram?
  – Moti
  Jul 17 at 5:49
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I recently came across the following problem in olympiad training material:

A circle $k$ is internally tangent to sides $AB, AC$ of $Delta ABC$ and its circumcircle in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of $XY$ and $AZ$. Prove that $measuredangle XTB=measuredangle CTY$.

By inverting through the circle with center $A$ that is passing through $T$, I managed to transform the problem into the following, in my opinion easier, version:

The $A$-excircle of $Delta ABC$ touches $AB, AC$ and $BC$ in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of the circumcircle of $Delta AXY$ with $AZ$. Prove $measuredangle ATB=measuredangle CTA$.

However, I have not been able to solve the second version of the problem either and only got as far as $measuredangle ATX=measuredangle YTA$ which leaves $measuredangle YTC=measuredangle BTX$ to prove.
Furthermore, the circle through $A, X $ and $Y $ is obviously Thales' circle with diameter $AM $ where $M $ is the center of the excircle.

Does anyone have a suggestion how to proceed or how to solve the initial problem without an inversion?

geometry contest-math circle triangle

share|cite|improve this question

edited Jul 17 at 8:40

asked Jul 16 at 20:18

mxian

563216

I recently came across the following problem in olympiad training material:

A circle $k$ is internally tangent to sides $AB, AC$ of $Delta ABC$ and its circumcircle in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of $XY$ and $AZ$. Prove that $measuredangle XTB=measuredangle CTY$.

By inverting through the circle with center $A$ that is passing through $T$, I managed to transform the problem into the following, in my opinion easier, version:

The $A$-excircle of $Delta ABC$ touches $AB, AC$ and $BC$ in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of the circumcircle of $Delta AXY$ with $AZ$. Prove $measuredangle ATB=measuredangle CTA$.

However, I have not been able to solve the second version of the problem either and only got as far as $measuredangle ATX=measuredangle YTA$ which leaves $measuredangle YTC=measuredangle BTX$ to prove.
Furthermore, the circle through $A, X $ and $Y $ is obviously Thales' circle with diameter $AM $ where $M $ is the center of the excircle.

Does anyone have a suggestion how to proceed or how to solve the initial problem without an inversion?

geometry contest-math circle triangle

share|cite|improve this question

edited Jul 17 at 8:40

asked Jul 16 at 20:18

mxian

563216

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 8:40

edited Jul 17 at 8:40

edited Jul 17 at 8:40

asked Jul 16 at 20:18

mxian

563216

asked Jul 16 at 20:18

mxian

563216

asked Jul 16 at 20:18

mxian

563216

563216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you add a diagram?
  – Moti
  Jul 17 at 5:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you add a diagram?
  – Moti
  Jul 17 at 5:49
  

  
  
  
  

Could you add a diagram?
– Moti
Jul 17 at 5:49

Could you add a diagram?
– Moti
Jul 17 at 5:49

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

I'll split the solution into several small steps.

Let $ZX$ and $ZY$ intersect the circumcircle of $ABC$ again at $P$ and $Q$, respectively.

---

It is enough to prove that $triangle BXT sim triangle CYT$.

---

Since $angle BXT = angle TYC$, it is enough to prove that $dfracBXCY = dfracXTYT$.

---

Note that $ZA$ is a symmedian in $triangle XYZ$. Therefore $dfracXTYT=dfracXZ^2YZ^2$.

---

Prove that $XY parallel PQ$. Conclude that $dfracXZYZ = dfracPZQZ$.

---

Note that $XY parallel PQ$ implies that $ZA$ is a symmedian in $triangle ZPQ$ as well. Conclude that $dfracZPZQ = dfracAPAQ$.

---

Prove that $P$ and $Q$ are the midpoints of arc $AB$ and $AC$, respectively.

---

Prove that $triangle ZBX sim triangle ZPA$ and $triangle ZCY sim triangle ZQA$. Conclude that $dfracBXXZ = dfracPAAZ$ and $dfracCYYZ=dfracQAAZ$.

---

Combine all ingredients.

share|cite|improve this answer

edited Jul 17 at 12:47

answered Jul 17 at 10:07

timon92

3,7141724

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Note that $ZA $ is a symmedian in $Delta XYZ $." Why is that? At least to me, that is not trivial. Could you add a more detailed explanation of that step?
  – mxian
  Jul 17 at 11:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a well-known property of symmedians. You can read a proof e.g. here: cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml
  – timon92
  Jul 17 at 11:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, that has made it clear. I'd suggest you just edit that writing error you made: It should be $dfrac BXXZ=dfrac APAZ $, otherwise it won't work out.
  – mxian
  Jul 17 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're absolutely right. I have corrected it.
  – timon92
  Jul 17 at 12:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

Proof

Lemma Let the circle $c_1$ internally contact the circle $c_2$ at $T$, and
contact the chord $AB$ of $c_2$ at $C$. Then $TC$ bisects $angle ATB$.

We are not going to give the proof, because it's very simple, if you notice that $c_1$ and $c_2$ are homothetic with the homothetic center $T$. Now, let's turn to prove the original statement.

By the lemma, we have $ZX$ bisects $angle AZB$ and $ZY$ bisects $angle AZC$. Thus $$fracBXXA=fracZBZA,~~~~~~fracCYYA=fracZCZA.$$
Since $XA=YA$, therefore, $$fracBXCY=fracZBZC=fracsinangle BAZsin angle CAZ=fracdfrac12cdot AXcdot AT cdot sinangle BAZ dfrac12 cdot AYcdot AT cdot sinangle CAZ=fracS_Delta AXTS_Delta AYT=fracXTYT.$$

Notice that $angle BXT=angle CYT$. Hence, $$Delta BXT sim Delta CYT.$$
It follows that $$angle XTB=angle YTC,$$ which is desired.

share|cite|improve this answer

answered Jul 17 at 16:33

mengdie1982

2,972216

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

I'll split the solution into several small steps.

Let $ZX$ and $ZY$ intersect the circumcircle of $ABC$ again at $P$ and $Q$, respectively.

---

It is enough to prove that $triangle BXT sim triangle CYT$.

---

Since $angle BXT = angle TYC$, it is enough to prove that $dfracBXCY = dfracXTYT$.

---

Note that $ZA$ is a symmedian in $triangle XYZ$. Therefore $dfracXTYT=dfracXZ^2YZ^2$.

---

Prove that $XY parallel PQ$. Conclude that $dfracXZYZ = dfracPZQZ$.

---

Note that $XY parallel PQ$ implies that $ZA$ is a symmedian in $triangle ZPQ$ as well. Conclude that $dfracZPZQ = dfracAPAQ$.

---

Prove that $P$ and $Q$ are the midpoints of arc $AB$ and $AC$, respectively.

---

Prove that $triangle ZBX sim triangle ZPA$ and $triangle ZCY sim triangle ZQA$. Conclude that $dfracBXXZ = dfracPAAZ$ and $dfracCYYZ=dfracQAAZ$.

---

Combine all ingredients.

share|cite|improve this answer

edited Jul 17 at 12:47

answered Jul 17 at 10:07

timon92

3,7141724

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Note that $ZA $ is a symmedian in $Delta XYZ $." Why is that? At least to me, that is not trivial. Could you add a more detailed explanation of that step?
  – mxian
  Jul 17 at 11:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a well-known property of symmedians. You can read a proof e.g. here: cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml
  – timon92
  Jul 17 at 11:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, that has made it clear. I'd suggest you just edit that writing error you made: It should be $dfrac BXXZ=dfrac APAZ $, otherwise it won't work out.
  – mxian
  Jul 17 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're absolutely right. I have corrected it.
  – timon92
  Jul 17 at 12:47
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

I'll split the solution into several small steps.

Let $ZX$ and $ZY$ intersect the circumcircle of $ABC$ again at $P$ and $Q$, respectively.

---

It is enough to prove that $triangle BXT sim triangle CYT$.

---

Since $angle BXT = angle TYC$, it is enough to prove that $dfracBXCY = dfracXTYT$.

---

Note that $ZA$ is a symmedian in $triangle XYZ$. Therefore $dfracXTYT=dfracXZ^2YZ^2$.

---

Prove that $XY parallel PQ$. Conclude that $dfracXZYZ = dfracPZQZ$.

---

Note that $XY parallel PQ$ implies that $ZA$ is a symmedian in $triangle ZPQ$ as well. Conclude that $dfracZPZQ = dfracAPAQ$.

---

Prove that $P$ and $Q$ are the midpoints of arc $AB$ and $AC$, respectively.

---

Prove that $triangle ZBX sim triangle ZPA$ and $triangle ZCY sim triangle ZQA$. Conclude that $dfracBXXZ = dfracPAAZ$ and $dfracCYYZ=dfracQAAZ$.

---

Combine all ingredients.

share|cite|improve this answer

edited Jul 17 at 12:47

answered Jul 17 at 10:07

timon92

3,7141724

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Note that $ZA $ is a symmedian in $Delta XYZ $." Why is that? At least to me, that is not trivial. Could you add a more detailed explanation of that step?
  – mxian
  Jul 17 at 11:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a well-known property of symmedians. You can read a proof e.g. here: cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml
  – timon92
  Jul 17 at 11:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, that has made it clear. I'd suggest you just edit that writing error you made: It should be $dfrac BXXZ=dfrac APAZ $, otherwise it won't work out.
  – mxian
  Jul 17 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're absolutely right. I have corrected it.
  – timon92
  Jul 17 at 12:47
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

I'll split the solution into several small steps.

Let $ZX$ and $ZY$ intersect the circumcircle of $ABC$ again at $P$ and $Q$, respectively.

---

It is enough to prove that $triangle BXT sim triangle CYT$.

---

Since $angle BXT = angle TYC$, it is enough to prove that $dfracBXCY = dfracXTYT$.

---

Note that $ZA$ is a symmedian in $triangle XYZ$. Therefore $dfracXTYT=dfracXZ^2YZ^2$.

---

Prove that $XY parallel PQ$. Conclude that $dfracXZYZ = dfracPZQZ$.

---

Note that $XY parallel PQ$ implies that $ZA$ is a symmedian in $triangle ZPQ$ as well. Conclude that $dfracZPZQ = dfracAPAQ$.

---

Prove that $P$ and $Q$ are the midpoints of arc $AB$ and $AC$, respectively.

---

Prove that $triangle ZBX sim triangle ZPA$ and $triangle ZCY sim triangle ZQA$. Conclude that $dfracBXXZ = dfracPAAZ$ and $dfracCYYZ=dfracQAAZ$.

---

Combine all ingredients.

share|cite|improve this answer

edited Jul 17 at 12:47

answered Jul 17 at 10:07

timon92

3,7141724

I'll split the solution into several small steps.

Let $ZX$ and $ZY$ intersect the circumcircle of $ABC$ again at $P$ and $Q$, respectively.

---

It is enough to prove that $triangle BXT sim triangle CYT$.

---

Since $angle BXT = angle TYC$, it is enough to prove that $dfracBXCY = dfracXTYT$.

---

Note that $ZA$ is a symmedian in $triangle XYZ$. Therefore $dfracXTYT=dfracXZ^2YZ^2$.

---

Prove that $XY parallel PQ$. Conclude that $dfracXZYZ = dfracPZQZ$.

---

Note that $XY parallel PQ$ implies that $ZA$ is a symmedian in $triangle ZPQ$ as well. Conclude that $dfracZPZQ = dfracAPAQ$.

---

Prove that $P$ and $Q$ are the midpoints of arc $AB$ and $AC$, respectively.

---

Prove that $triangle ZBX sim triangle ZPA$ and $triangle ZCY sim triangle ZQA$. Conclude that $dfracBXXZ = dfracPAAZ$ and $dfracCYYZ=dfracQAAZ$.

---

Combine all ingredients.

share|cite|improve this answer

edited Jul 17 at 12:47

answered Jul 17 at 10:07

timon92

3,7141724

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 12:47

edited Jul 17 at 12:47

edited Jul 17 at 12:47

answered Jul 17 at 10:07

timon92

3,7141724

answered Jul 17 at 10:07

timon92

3,7141724

answered Jul 17 at 10:07

timon92

3,7141724

3,7141724

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Note that $ZA $ is a symmedian in $Delta XYZ $." Why is that? At least to me, that is not trivial. Could you add a more detailed explanation of that step?
  – mxian
  Jul 17 at 11:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a well-known property of symmedians. You can read a proof e.g. here: cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml
  – timon92
  Jul 17 at 11:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, that has made it clear. I'd suggest you just edit that writing error you made: It should be $dfrac BXXZ=dfrac APAZ $, otherwise it won't work out.
  – mxian
  Jul 17 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're absolutely right. I have corrected it.
  – timon92
  Jul 17 at 12:47
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "Note that $ZA $ is a symmedian in $Delta XYZ $." Why is that? At least to me, that is not trivial. Could you add a more detailed explanation of that step?
  – mxian
  Jul 17 at 11:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a well-known property of symmedians. You can read a proof e.g. here: cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml
  – timon92
  Jul 17 at 11:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much, that has made it clear. I'd suggest you just edit that writing error you made: It should be $dfrac BXXZ=dfrac APAZ $, otherwise it won't work out.
  – mxian
  Jul 17 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're absolutely right. I have corrected it.
  – timon92
  Jul 17 at 12:47
  

  
  
  
  

"Note that $ZA $ is a symmedian in $Delta XYZ $." Why is that? At least to me, that is not trivial. Could you add a more detailed explanation of that step?
– mxian
Jul 17 at 11:25

"Note that $ZA $ is a symmedian in $Delta XYZ $." Why is that? At least to me, that is not trivial. Could you add a more detailed explanation of that step?
– mxian
Jul 17 at 11:25

This is a well-known property of symmedians. You can read a proof e.g. here: cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml
– timon92
Jul 17 at 11:37

This is a well-known property of symmedians. You can read a proof e.g. here: cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml
– timon92
Jul 17 at 11:37

Thank you very much, that has made it clear. I'd suggest you just edit that writing error you made: It should be $dfrac BXXZ=dfrac APAZ $, otherwise it won't work out.
– mxian
Jul 17 at 12:18

Thank you very much, that has made it clear. I'd suggest you just edit that writing error you made: It should be $dfrac BXXZ=dfrac APAZ $, otherwise it won't work out.
– mxian
Jul 17 at 12:18

You're absolutely right. I have corrected it.
– timon92
Jul 17 at 12:47

You're absolutely right. I have corrected it.
– timon92
Jul 17 at 12:47

add a comment | 

up vote
1
down vote

Proof

Lemma Let the circle $c_1$ internally contact the circle $c_2$ at $T$, and
contact the chord $AB$ of $c_2$ at $C$. Then $TC$ bisects $angle ATB$.

We are not going to give the proof, because it's very simple, if you notice that $c_1$ and $c_2$ are homothetic with the homothetic center $T$. Now, let's turn to prove the original statement.

By the lemma, we have $ZX$ bisects $angle AZB$ and $ZY$ bisects $angle AZC$. Thus $$fracBXXA=fracZBZA,~~~~~~fracCYYA=fracZCZA.$$
Since $XA=YA$, therefore, $$fracBXCY=fracZBZC=fracsinangle BAZsin angle CAZ=fracdfrac12cdot AXcdot AT cdot sinangle BAZ dfrac12 cdot AYcdot AT cdot sinangle CAZ=fracS_Delta AXTS_Delta AYT=fracXTYT.$$

Notice that $angle BXT=angle CYT$. Hence, $$Delta BXT sim Delta CYT.$$
It follows that $$angle XTB=angle YTC,$$ which is desired.

share|cite|improve this answer

answered Jul 17 at 16:33

mengdie1982

2,972216

add a comment | 

up vote
1
down vote

Proof

Lemma Let the circle $c_1$ internally contact the circle $c_2$ at $T$, and
contact the chord $AB$ of $c_2$ at $C$. Then $TC$ bisects $angle ATB$.

We are not going to give the proof, because it's very simple, if you notice that $c_1$ and $c_2$ are homothetic with the homothetic center $T$. Now, let's turn to prove the original statement.

By the lemma, we have $ZX$ bisects $angle AZB$ and $ZY$ bisects $angle AZC$. Thus $$fracBXXA=fracZBZA,~~~~~~fracCYYA=fracZCZA.$$
Since $XA=YA$, therefore, $$fracBXCY=fracZBZC=fracsinangle BAZsin angle CAZ=fracdfrac12cdot AXcdot AT cdot sinangle BAZ dfrac12 cdot AYcdot AT cdot sinangle CAZ=fracS_Delta AXTS_Delta AYT=fracXTYT.$$

Notice that $angle BXT=angle CYT$. Hence, $$Delta BXT sim Delta CYT.$$
It follows that $$angle XTB=angle YTC,$$ which is desired.

share|cite|improve this answer

answered Jul 17 at 16:33

mengdie1982

2,972216

add a comment | 

up vote
1
down vote

up vote
1
down vote

Proof

Lemma Let the circle $c_1$ internally contact the circle $c_2$ at $T$, and
contact the chord $AB$ of $c_2$ at $C$. Then $TC$ bisects $angle ATB$.

We are not going to give the proof, because it's very simple, if you notice that $c_1$ and $c_2$ are homothetic with the homothetic center $T$. Now, let's turn to prove the original statement.

By the lemma, we have $ZX$ bisects $angle AZB$ and $ZY$ bisects $angle AZC$. Thus $$fracBXXA=fracZBZA,~~~~~~fracCYYA=fracZCZA.$$
Since $XA=YA$, therefore, $$fracBXCY=fracZBZC=fracsinangle BAZsin angle CAZ=fracdfrac12cdot AXcdot AT cdot sinangle BAZ dfrac12 cdot AYcdot AT cdot sinangle CAZ=fracS_Delta AXTS_Delta AYT=fracXTYT.$$

Notice that $angle BXT=angle CYT$. Hence, $$Delta BXT sim Delta CYT.$$
It follows that $$angle XTB=angle YTC,$$ which is desired.

share|cite|improve this answer

answered Jul 17 at 16:33

mengdie1982

2,972216

Proof

Lemma Let the circle $c_1$ internally contact the circle $c_2$ at $T$, and
contact the chord $AB$ of $c_2$ at $C$. Then $TC$ bisects $angle ATB$.

We are not going to give the proof, because it's very simple, if you notice that $c_1$ and $c_2$ are homothetic with the homothetic center $T$. Now, let's turn to prove the original statement.

By the lemma, we have $ZX$ bisects $angle AZB$ and $ZY$ bisects $angle AZC$. Thus $$fracBXXA=fracZBZA,~~~~~~fracCYYA=fracZCZA.$$
Since $XA=YA$, therefore, $$fracBXCY=fracZBZC=fracsinangle BAZsin angle CAZ=fracdfrac12cdot AXcdot AT cdot sinangle BAZ dfrac12 cdot AYcdot AT cdot sinangle CAZ=fracS_Delta AXTS_Delta AYT=fracXTYT.$$

Notice that $angle BXT=angle CYT$. Hence, $$Delta BXT sim Delta CYT.$$
It follows that $$angle XTB=angle YTC,$$ which is desired.

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answered Jul 17 at 16:33

mengdie1982

2,972216

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share|cite|improve this answer

answered Jul 17 at 16:33

mengdie1982

2,972216

answered Jul 17 at 16:33

mengdie1982

2,972216

answered Jul 17 at 16:33

mengdie1982

2,972216

2,972216

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