Prove that $BH=AH$
up vote
5
down vote
favorite
A triangle $ABC$ is given. There's a point $P$ inside it and also it is connected to point $H$, which lies on edge $BC$ ($H$ must not be the middle point of edge $BC$). Turns out, that bisector of angle $∠AHP$ is perpendicular to edge $BC$. Also, $BP=AC$ and $∠PCH=∠ABC$. Prove that $BH=AH$.
What I found out was that since triangle $ACH$ hypothetically should be equal to triangle to $PBH$, we could prove what we need by proving that for example $PH=CH$ - though again, I have no idea how to do it. Could you recommend me any smart lines or segments to draw?
geometry trigonometry euclidean-geometry triangle problem-solving
edited Dec 3 at 14:04
Zvi
3,875328
asked Dec 2 at 20:06
Severus156
1707
add a comment |
up vote
5
down vote
favorite
A triangle $ABC$ is given. There's a point $P$ inside it and also it is connected to point $H$, which lies on edge $BC$ ($H$ must not be the middle point of edge $BC$). Turns out, that bisector of angle $∠AHP$ is perpendicular to edge $BC$. Also, $BP=AC$ and $∠PCH=∠ABC$. Prove that $BH=AH$.
What I found out was that since triangle $ACH$ hypothetically should be equal to triangle to $PBH$, we could prove what we need by proving that for example $PH=CH$ - though again, I have no idea how to do it. Could you recommend me any smart lines or segments to draw?
geometry trigonometry euclidean-geometry triangle problem-solving
edited Dec 3 at 14:04
Zvi
3,875328
asked Dec 2 at 20:06
Severus156
1707
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
A triangle $ABC$ is given. There's a point $P$ inside it and also it is connected to point $H$, which lies on edge $BC$ ($H$ must not be the middle point of edge $BC$). Turns out, that bisector of angle $∠AHP$ is perpendicular to edge $BC$. Also, $BP=AC$ and $∠PCH=∠ABC$. Prove that $BH=AH$.
What I found out was that since triangle $ACH$ hypothetically should be equal to triangle to $PBH$, we could prove what we need by proving that for example $PH=CH$ - though again, I have no idea how to do it. Could you recommend me any smart lines or segments to draw?
geometry trigonometry euclidean-geometry triangle problem-solving
edited Dec 3 at 14:04
Zvi
3,875328
asked Dec 2 at 20:06
Severus156
1707
A triangle $ABC$ is given. There's a point $P$ inside it and also it is connected to point $H$, which lies on edge $BC$ ($H$ must not be the middle point of edge $BC$). Turns out, that bisector of angle $∠AHP$ is perpendicular to edge $BC$. Also, $BP=AC$ and $∠PCH=∠ABC$. Prove that $BH=AH$.
What I found out was that since triangle $ACH$ hypothetically should be equal to triangle to $PBH$, we could prove what we need by proving that for example $PH=CH$ - though again, I have no idea how to do it. Could you recommend me any smart lines or segments to draw?
geometry trigonometry euclidean-geometry triangle problem-solving
geometry trigonometry euclidean-geometry triangle problem-solving
edited Dec 3 at 14:04
Zvi
3,875328
asked Dec 2 at 20:06
Severus156
1707
edited Dec 3 at 14:04
Zvi
3,875328
asked Dec 2 at 20:06
Severus156
1707
edited Dec 3 at 14:04
Zvi
3,875328
edited Dec 3 at 14:04
Zvi
3,875328
edited Dec 3 at 14:04
Zvi
3,875328
3,875328
asked Dec 2 at 20:06
Severus156
1707
asked Dec 2 at 20:06
Severus156
1707
asked Dec 2 at 20:06
Severus156
1707
1707
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Since the angle bisector of $angle AHP$ is perpendicular to $BC$, $angle PHC = angle AHB$. This and $angle PCH = angle HBA$ imply that $triangle PHC$ and $triangle AHB$ are similar. Thus
$$frac{PH}{CH} = frac{AH}{BH},$$
or
$$PHcdot BH = AH cdot CH.$$
From this and $BP=AC, angle BHP=angle CHA$, the cosine law in $triangle PBH$ and $triangle CAH$ would give
$$PH^2 + BH^2 = AH^2 + CH^2.$$
Thus $(PH,BH) = (AH, CH)$ or $(PH,BH) = (CH,AH)$. But $H$ is not the midpoint of $BC$, i.e. $BHneq CH$, so we are done.
answered Dec 3 at 4:02
Quang Hoang
12.2k1131
add a comment |
up vote
1
down vote
Since the angle bisector of $angle , AHP$ is perpendicular to $BC$, then $BC$ is in fact the outer angle bisector of $angle , AHP$ so $angle , AHC = angle , PHB$.
Let point $A'$ be the reflection image of point $A$ with respect to the line $BC$. Then triangle $Delta , A'BC$ is the reflection image of triangle $Delta, ABC$ with respect to $BC$. Consequently,
$$angle , A'HC = angle , AHC = angle, PHB$$
which is possible if and only if the points $P, H$ and $A'$ are collinear, i.e. $H in PA'$.
By assumption, $$angle , PCB = angle , BCB = angle , ABC$$ But by construction,
$$angle , ABC = angle , A'BC$$ which yields
$$angle , PCB = angle , A'BC$$
and therefore the segments $$CP , || , A'B$$ i.e. the quad $CPBA'$ is a trapezoid. By assumption, however, $BP = AC$ and by construction (reflection symmetry) $AC = A'C$ so
$$BP = AC$$ Therefore the trapezoid $CPBA'$ is isosceles, which implies that $A'H = BH$. But since $A'$ is the symmetric image of $A$ with respect to $BC$
$$AH = A'H = BH$$
answered Dec 3 at 17:49
Futurologist
7,2962519
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since the angle bisector of $angle AHP$ is perpendicular to $BC$, $angle PHC = angle AHB$. This and $angle PCH = angle HBA$ imply that $triangle PHC$ and $triangle AHB$ are similar. Thus
$$frac{PH}{CH} = frac{AH}{BH},$$
or
$$PHcdot BH = AH cdot CH.$$
From this and $BP=AC, angle BHP=angle CHA$, the cosine law in $triangle PBH$ and $triangle CAH$ would give
$$PH^2 + BH^2 = AH^2 + CH^2.$$
Thus $(PH,BH) = (AH, CH)$ or $(PH,BH) = (CH,AH)$. But $H$ is not the midpoint of $BC$, i.e. $BHneq CH$, so we are done.
answered Dec 3 at 4:02
Quang Hoang
12.2k1131
add a comment |
up vote
1
down vote
accepted
Since the angle bisector of $angle AHP$ is perpendicular to $BC$, $angle PHC = angle AHB$. This and $angle PCH = angle HBA$ imply that $triangle PHC$ and $triangle AHB$ are similar. Thus
$$frac{PH}{CH} = frac{AH}{BH},$$
or
$$PHcdot BH = AH cdot CH.$$
From this and $BP=AC, angle BHP=angle CHA$, the cosine law in $triangle PBH$ and $triangle CAH$ would give
$$PH^2 + BH^2 = AH^2 + CH^2.$$
Thus $(PH,BH) = (AH, CH)$ or $(PH,BH) = (CH,AH)$. But $H$ is not the midpoint of $BC$, i.e. $BHneq CH$, so we are done.
answered Dec 3 at 4:02
Quang Hoang
12.2k1131
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since the angle bisector of $angle AHP$ is perpendicular to $BC$, $angle PHC = angle AHB$. This and $angle PCH = angle HBA$ imply that $triangle PHC$ and $triangle AHB$ are similar. Thus
$$frac{PH}{CH} = frac{AH}{BH},$$
or
$$PHcdot BH = AH cdot CH.$$
From this and $BP=AC, angle BHP=angle CHA$, the cosine law in $triangle PBH$ and $triangle CAH$ would give
$$PH^2 + BH^2 = AH^2 + CH^2.$$
Thus $(PH,BH) = (AH, CH)$ or $(PH,BH) = (CH,AH)$. But $H$ is not the midpoint of $BC$, i.e. $BHneq CH$, so we are done.
answered Dec 3 at 4:02
Quang Hoang
12.2k1131
Since the angle bisector of $angle AHP$ is perpendicular to $BC$, $angle PHC = angle AHB$. This and $angle PCH = angle HBA$ imply that $triangle PHC$ and $triangle AHB$ are similar. Thus
$$frac{PH}{CH} = frac{AH}{BH},$$
or
$$PHcdot BH = AH cdot CH.$$
From this and $BP=AC, angle BHP=angle CHA$, the cosine law in $triangle PBH$ and $triangle CAH$ would give
$$PH^2 + BH^2 = AH^2 + CH^2.$$
Thus $(PH,BH) = (AH, CH)$ or $(PH,BH) = (CH,AH)$. But $H$ is not the midpoint of $BC$, i.e. $BHneq CH$, so we are done.
answered Dec 3 at 4:02
Quang Hoang
12.2k1131
answered Dec 3 at 4:02
Quang Hoang
12.2k1131
answered Dec 3 at 4:02
Quang Hoang
12.2k1131
answered Dec 3 at 4:02
Quang Hoang
12.2k1131
12.2k1131
add a comment |
add a comment |
up vote
1
down vote
Since the angle bisector of $angle , AHP$ is perpendicular to $BC$, then $BC$ is in fact the outer angle bisector of $angle , AHP$ so $angle , AHC = angle , PHB$.
Let point $A'$ be the reflection image of point $A$ with respect to the line $BC$. Then triangle $Delta , A'BC$ is the reflection image of triangle $Delta, ABC$ with respect to $BC$. Consequently,
$$angle , A'HC = angle , AHC = angle, PHB$$
which is possible if and only if the points $P, H$ and $A'$ are collinear, i.e. $H in PA'$.
By assumption, $$angle , PCB = angle , BCB = angle , ABC$$ But by construction,
$$angle , ABC = angle , A'BC$$ which yields
$$angle , PCB = angle , A'BC$$
and therefore the segments $$CP , || , A'B$$ i.e. the quad $CPBA'$ is a trapezoid. By assumption, however, $BP = AC$ and by construction (reflection symmetry) $AC = A'C$ so
$$BP = AC$$ Therefore the trapezoid $CPBA'$ is isosceles, which implies that $A'H = BH$. But since $A'$ is the symmetric image of $A$ with respect to $BC$
$$AH = A'H = BH$$
answered Dec 3 at 17:49
Futurologist
7,2962519
add a comment |
up vote
1
down vote
Since the angle bisector of $angle , AHP$ is perpendicular to $BC$, then $BC$ is in fact the outer angle bisector of $angle , AHP$ so $angle , AHC = angle , PHB$.
Let point $A'$ be the reflection image of point $A$ with respect to the line $BC$. Then triangle $Delta , A'BC$ is the reflection image of triangle $Delta, ABC$ with respect to $BC$. Consequently,
$$angle , A'HC = angle , AHC = angle, PHB$$
which is possible if and only if the points $P, H$ and $A'$ are collinear, i.e. $H in PA'$.
By assumption, $$angle , PCB = angle , BCB = angle , ABC$$ But by construction,
$$angle , ABC = angle , A'BC$$ which yields
$$angle , PCB = angle , A'BC$$
and therefore the segments $$CP , || , A'B$$ i.e. the quad $CPBA'$ is a trapezoid. By assumption, however, $BP = AC$ and by construction (reflection symmetry) $AC = A'C$ so
$$BP = AC$$ Therefore the trapezoid $CPBA'$ is isosceles, which implies that $A'H = BH$. But since $A'$ is the symmetric image of $A$ with respect to $BC$
$$AH = A'H = BH$$
answered Dec 3 at 17:49
Futurologist
7,2962519
add a comment |
up vote
1
down vote
up vote
1
down vote
Since the angle bisector of $angle , AHP$ is perpendicular to $BC$, then $BC$ is in fact the outer angle bisector of $angle , AHP$ so $angle , AHC = angle , PHB$.
Let point $A'$ be the reflection image of point $A$ with respect to the line $BC$. Then triangle $Delta , A'BC$ is the reflection image of triangle $Delta, ABC$ with respect to $BC$. Consequently,
$$angle , A'HC = angle , AHC = angle, PHB$$
which is possible if and only if the points $P, H$ and $A'$ are collinear, i.e. $H in PA'$.
By assumption, $$angle , PCB = angle , BCB = angle , ABC$$ But by construction,
$$angle , ABC = angle , A'BC$$ which yields
$$angle , PCB = angle , A'BC$$
and therefore the segments $$CP , || , A'B$$ i.e. the quad $CPBA'$ is a trapezoid. By assumption, however, $BP = AC$ and by construction (reflection symmetry) $AC = A'C$ so
$$BP = AC$$ Therefore the trapezoid $CPBA'$ is isosceles, which implies that $A'H = BH$. But since $A'$ is the symmetric image of $A$ with respect to $BC$
$$AH = A'H = BH$$
answered Dec 3 at 17:49
Futurologist
7,2962519
Since the angle bisector of $angle , AHP$ is perpendicular to $BC$, then $BC$ is in fact the outer angle bisector of $angle , AHP$ so $angle , AHC = angle , PHB$.
Let point $A'$ be the reflection image of point $A$ with respect to the line $BC$. Then triangle $Delta , A'BC$ is the reflection image of triangle $Delta, ABC$ with respect to $BC$. Consequently,
$$angle , A'HC = angle , AHC = angle, PHB$$
which is possible if and only if the points $P, H$ and $A'$ are collinear, i.e. $H in PA'$.
By assumption, $$angle , PCB = angle , BCB = angle , ABC$$ But by construction,
$$angle , ABC = angle , A'BC$$ which yields
$$angle , PCB = angle , A'BC$$
and therefore the segments $$CP , || , A'B$$ i.e. the quad $CPBA'$ is a trapezoid. By assumption, however, $BP = AC$ and by construction (reflection symmetry) $AC = A'C$ so
$$BP = AC$$ Therefore the trapezoid $CPBA'$ is isosceles, which implies that $A'H = BH$. But since $A'$ is the symmetric image of $A$ with respect to $BC$
$$AH = A'H = BH$$
answered Dec 3 at 17:49
Futurologist
7,2962519
answered Dec 3 at 17:49
Futurologist
7,2962519
answered Dec 3 at 17:49
Futurologist
7,2962519
answered Dec 3 at 17:49
Futurologist
7,2962519
7,2962519
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023110%2fprove-that-bh-ah%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
