AD and BE are the altitudes of the triangle ABC with orthocentre H,which lies in the interior of the...
$begingroup$
After forming some equations we get angle DAC=EBC .After this I hit a dead end but I think this would require cosine law or some other trignometric relation.Please help (HAVE BEEN WORKING ON THIS FOR A LONG TIME)
geometry
asked Dec 12 '18 at 16:26
mtommtom
254
$endgroup$
|
show 4 more comments
$begingroup$
After forming some equations we get angle DAC=EBC .After this I hit a dead end but I think this would require cosine law or some other trignometric relation.Please help (HAVE BEEN WORKING ON THIS FOR A LONG TIME)
geometry
asked Dec 12 '18 at 16:26
mtommtom
254
$endgroup$
$begingroup$
Maybe provide a picture to reason about?
$endgroup$
– Federico
Dec 12 '18 at 16:34
$begingroup$
i am new to this site so how do you do that?
$endgroup$
– mtom
Dec 12 '18 at 16:35
$begingroup$
You may use the geogebra app
$endgroup$
– Fareed AF
Dec 12 '18 at 16:54
$begingroup$
i made a figure on paint
$endgroup$
– mtom
Dec 12 '18 at 17:02
$begingroup$
MS paint is not easy to work with so forgive the diagram
$endgroup$
– mtom
Dec 12 '18 at 17:03
|
show 4 more comments
$begingroup$
After forming some equations we get angle DAC=EBC .After this I hit a dead end but I think this would require cosine law or some other trignometric relation.Please help (HAVE BEEN WORKING ON THIS FOR A LONG TIME)
geometry
asked Dec 12 '18 at 16:26
mtommtom
254
$endgroup$
After forming some equations we get angle DAC=EBC .After this I hit a dead end but I think this would require cosine law or some other trignometric relation.Please help (HAVE BEEN WORKING ON THIS FOR A LONG TIME)
geometry
geometry
asked Dec 12 '18 at 16:26
mtommtom
254
asked Dec 12 '18 at 16:26
mtommtom
254
edited Dec 12 '18 at 17:00
mtom
asked Dec 12 '18 at 16:26
mtommtom
254
asked Dec 12 '18 at 16:26
mtommtom
254
asked Dec 12 '18 at 16:26
mtommtom
254
254
$begingroup$
Maybe provide a picture to reason about?
$endgroup$
– Federico
Dec 12 '18 at 16:34
$begingroup$
i am new to this site so how do you do that?
$endgroup$
– mtom
Dec 12 '18 at 16:35
$begingroup$
You may use the geogebra app
$endgroup$
– Fareed AF
Dec 12 '18 at 16:54
$begingroup$
i made a figure on paint
$endgroup$
– mtom
Dec 12 '18 at 17:02
$begingroup$
MS paint is not easy to work with so forgive the diagram
$endgroup$
– mtom
Dec 12 '18 at 17:03
|
show 4 more comments
$begingroup$
Maybe provide a picture to reason about?
$endgroup$
– Federico
Dec 12 '18 at 16:34
$begingroup$
i am new to this site so how do you do that?
$endgroup$
– mtom
Dec 12 '18 at 16:35
$begingroup$
You may use the geogebra app
$endgroup$
– Fareed AF
Dec 12 '18 at 16:54
$begingroup$
i made a figure on paint
$endgroup$
– mtom
Dec 12 '18 at 17:02
$begingroup$
MS paint is not easy to work with so forgive the diagram
$endgroup$
– mtom
Dec 12 '18 at 17:03
$begingroup$
Maybe provide a picture to reason about?
$endgroup$
– Federico
Dec 12 '18 at 16:34
$begingroup$
Maybe provide a picture to reason about?
$endgroup$
– Federico
Dec 12 '18 at 16:34
$begingroup$
i am new to this site so how do you do that?
$endgroup$
– mtom
Dec 12 '18 at 16:35
$begingroup$
i am new to this site so how do you do that?
$endgroup$
– mtom
Dec 12 '18 at 16:35
$begingroup$
You may use the geogebra app
$endgroup$
– Fareed AF
Dec 12 '18 at 16:54
$begingroup$
You may use the geogebra app
$endgroup$
– Fareed AF
Dec 12 '18 at 16:54
$begingroup$
i made a figure on paint
$endgroup$
– mtom
Dec 12 '18 at 17:02
$begingroup$
i made a figure on paint
$endgroup$
– mtom
Dec 12 '18 at 17:02
$begingroup$
MS paint is not easy to work with so forgive the diagram
$endgroup$
– mtom
Dec 12 '18 at 17:03
$begingroup$
MS paint is not easy to work with so forgive the diagram
$endgroup$
– mtom
Dec 12 '18 at 17:03
|
show 4 more comments
1 Answer
1
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oldest
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$begingroup$
Since $angle HBD = angle CAD$ and $BH=AC$, the two right triangles $triangle BHD$ and $triangle ACD$ are congruent. Thus $HD = CD$, or $angle HCD=45^circ$. Now $CHperp AB$, so $angle B + angle HCD = 90^circ$ and we have $angle B = 45^circ$.
answered Dec 12 '18 at 17:05
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $angle HBD = angle CAD$ and $BH=AC$, the two right triangles $triangle BHD$ and $triangle ACD$ are congruent. Thus $HD = CD$, or $angle HCD=45^circ$. Now $CHperp AB$, so $angle B + angle HCD = 90^circ$ and we have $angle B = 45^circ$.
answered Dec 12 '18 at 17:05
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
$begingroup$
Since $angle HBD = angle CAD$ and $BH=AC$, the two right triangles $triangle BHD$ and $triangle ACD$ are congruent. Thus $HD = CD$, or $angle HCD=45^circ$. Now $CHperp AB$, so $angle B + angle HCD = 90^circ$ and we have $angle B = 45^circ$.
answered Dec 12 '18 at 17:05
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
$begingroup$
Since $angle HBD = angle CAD$ and $BH=AC$, the two right triangles $triangle BHD$ and $triangle ACD$ are congruent. Thus $HD = CD$, or $angle HCD=45^circ$. Now $CHperp AB$, so $angle B + angle HCD = 90^circ$ and we have $angle B = 45^circ$.
answered Dec 12 '18 at 17:05
Quang HoangQuang Hoang
13.2k1233
$endgroup$
Since $angle HBD = angle CAD$ and $BH=AC$, the two right triangles $triangle BHD$ and $triangle ACD$ are congruent. Thus $HD = CD$, or $angle HCD=45^circ$. Now $CHperp AB$, so $angle B + angle HCD = 90^circ$ and we have $angle B = 45^circ$.
answered Dec 12 '18 at 17:05
Quang HoangQuang Hoang
13.2k1233
answered Dec 12 '18 at 17:05
Quang HoangQuang Hoang
13.2k1233
answered Dec 12 '18 at 17:05
Quang HoangQuang Hoang
13.2k1233
answered Dec 12 '18 at 17:05
Quang HoangQuang Hoang
13.2k1233
13.2k1233
add a comment |
add a comment |
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$begingroup$
Maybe provide a picture to reason about?
$endgroup$
– Federico
Dec 12 '18 at 16:34
$begingroup$
i am new to this site so how do you do that?
$endgroup$
– mtom
Dec 12 '18 at 16:35
$begingroup$
You may use the geogebra app
$endgroup$
– Fareed AF
Dec 12 '18 at 16:54
$begingroup$
i made a figure on paint
$endgroup$
– mtom
Dec 12 '18 at 17:02
$begingroup$
MS paint is not easy to work with so forgive the diagram
$endgroup$
– mtom
Dec 12 '18 at 17:03