It is known that in the oblique triangle ABC, the straight line with high BD and CE intersects at h, and the angle a = 45 degrees, so the degree of the angle BHC can be calculated? No picture, if you can,

∵△ ABC is an oblique triangle,
The ABC may be an acute triangle or an obtuse triangle,
When △ ABC is an acute triangle (as shown in Figure 1),
∵ BD and CE are the heights of △ ABC, ∠ a = 45 °,
∴∠ADB=∠BEH=90°,
∴∠ABD=90°-45°=45°,
∴∠BHC=∠ABH+∠BEH=45°+90°=135°.
When △ ABC is an obtuse triangle (as shown in Figure 2),
H is the intersection of two straight lines with the height of △ ABC, ∠ a = 45 °,
∴∠ABD=90°-45°=45°,
In RT △ EBH, ∠ BHC = 90 ° - abd = 90 ° - 45 ° = 45 °
In conclusion, the degree of ∠ BHC is 135 ° or 45 °

As shown in the figure, in △ ABC, ∠ A: ∠ ABC: ∠ ACB = 3:4:5, BD and CE are the heights of edges AC and ab respectively, BD and CE intersect at h, and the degree of ∠ BHC is calculated

Because ∠ A: ∠ ABC: ∠ ACB = 3:4:5, so ∠ a = 45 degrees
Because BD and CE are the heights of sides AC and AB, so ∠ AEH = ∠ ADH = 90 degrees
Therefore, EHD = 360-90 * 2-45 = 135 degrees, so BHC = ∠ EHD = 135 degrees

It is known that in non RT △ ABC, a = 45 ° and the straight line of high BD and CE intersects at point h. draw a graph and find out the degree of BHC

① As shown in Figure 1, when △ ABC is an acute triangle, ∵ BD and CE are the high lines of △ ABC, ∵ ADB = 90 °, ∵ BEC = 90 °, in △ abd, ∵ a = 45 °, ∵ abd = 90 ° - 45 ° = 45 °, ∵ BHC = ∵ abd + ∵ BEC = 45 ° + 90 ° = 135 °; when △ ABC is an obtuse triangle, ∵ BD and CE are △ ab

It is known that, as shown in the figure, in △ ABC, ∠ ABC = 66 ° and ∠ ACB = 54 ° be and CF are the heights of AC and AB on both sides, and they intersect at point h. calculate the degree of ∠ Abe and ∠ BHC

In ∵ △ ABC, ∵ ABC = 66 °, ∵ ACB = 54 °, ∵ a = 180 ° - ∵ ABC - ∵ ACB = 180 ° - 66 ° - 54 ° = 60 °, ∵ be ⊥ AC, ∵ AEB = 90 °, ∵ Abe = 90 ° - a = 90 ° - 60 ° = 30 °; similarly, ∵ CF ⊥ AB, ∵ BFC = 90 °, ∵ BHF = 90 ° - Abe = 90 ° - 30 ° = 60 °, ∵ BHC = 180 ° - BHF = 180 ° - 60 ° = 120 °

In the triangle ABC, we know that the angle ABC is 66 degrees, the angle ACB is 54 degrees, be is the height on the edge AC, CF is the height on the edge AB, h is the intersection of be and CF, and we can find the angle
BHC is solved by two methods

∵ be is the height on the AC side, so ∠ AEB = 90 °, ∵ Abe = 180 ° - BAC - ∠ AEB = 180 ° - 90 ° - 60 ° = 30 °. Similarly, ∠ ACF = 30 °, ∵ BHC = ∠ BEC + ∠ ACF = 90 ° + 30 ° = 12

It is known that, as shown in the figure, in △ ABC, ∠ ABC = 66 ° and ∠ ACB = 54 ° be and CF are the heights of AC and AB on both sides, and they intersect at point h. calculate the degree of ∠ Abe and ∠ BHC

In ∵ △ ABC, ∵ ABC = 66 °, ∵ ACB = 54 °, ∵ a = 180 ° - ∵ ABC - ∵ ACB = 180 ° - 66 ° - 54 ° = 60 °, ∵ be ⊥ AC, ∵ AEB = 90 °, ∵ Abe = 90 ° - a = 90 ° - 60 ° = 30 °; similarly, ∵ CF ⊥ AB, ∵ BFC = 90 °, ∵ BHF = 90 ° - Abe = 90 ° - 30 ° = 60

It is known that, as shown in the figure, in △ ABC, ∠ ABC = 66 ° and ∠ ACB = 54 ° be and CF are the heights of AC and AB on both sides, and they intersect at point h. calculate the degree of ∠ Abe and ∠ BHC

As shown in the figure, D is a point on the bisector of an outer angle of the triangle ABC. Prove that ab + AC is less than DB + DC

It is proved that △ CAD ≌ had can be easily proved by taking a point h on the extension line of Ba, making ah = AC, connecting DH
So CD = DH
In △ BDH, DH + DB > Hb
And DH = CD, ah = AC
∴DB+DC>AB+AC
Hope to help you

In the triangle ABC, ad is the bisector of the outer angle fac of angle a, and point D is on the extension line of line BC

It is proved that the parallel line of AB through C intersects AD and E
Then there is: ∠ AEC = ∠ fad, DB / DC = AB / EC
Because ad is the angular bisector of ∠ fac, that is ∠ fad = ∠ CAD
Therefore, AEC = CAD
So: AC = EC
So dB / DC = AB / AC

As shown in the figure, ∠ B = 90 ° in △ ABC, ab = BC, ad is the angular bisector of △ ABC, if BD = 1, then DC=______ ．

Make de ⊥ AC through D, intersect AC at point E, ∵ ad is the bisector of ∠ BAC, and ⊥ B = 90 °, that is, DB ⊥ AB, de ⊥ AC, ≁ DB = De, BD = 1, ≁ de = 1, and ≁ B = 90 °, ab = BC, ∩ ABC is isosceles right triangle, ∧ BAC = ∧ C = 45 °, and ∧ Dec = 90 °, ∧ Dec is isosceles right triangle, ∧ de = EC = 1. In RT △ Dec, according to Pythagorean theorem, DC = de2 + EC2 = 2 2

It is known that in the oblique triangle ABC, the straight line with high BD and CE intersects at h, and the angle a = 45 degrees, so the degree of the angle BHC can be calculated? No picture, if you can,