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In △ ABC, ab = BC, ∠ B = 90 °, M is the midpoint of AC, D.E is the moving point on AB and BC respectively, and BD = CE, the shape of △ DEM is obtained and proved
Delta DEM is an isosceles right triangle
prove:
Connect BM
∵ △ ABC is an isosceles right triangle and M is the midpoint of AC
∴BM⊥AC,MB=MC,∠B=∠MBD=45°
∵BD=CE
∴△BMD≌△CME
∴MD=ME,∠BMD=∠CME
∵∠BME+∠CME=90°
∴∠BME+∠BMD=90°
∴∠DME=90°
The △ DEM is an isosceles right triangle
As shown in the figure, in △ ABC, ∠ B = 90 °, ab = BC, BD = CE, M is the midpoint of AC edge, Verification: △ DEM is isosceles triangle
Proof: connect BM,
Because AB = BC, am = MC,
So BM ⊥ AC, and ∠ ABM = ∠ CBM = 1
2∠ABC=45°，
Because AB = BC,
So ∠ a = ∠ C = 180 ° − ABC
2=45°，
So ∠ a = ∠ ABM, so am = BM,
Because BD = CE, ab = BC, so ab-bd = bc-ce, that is, ad = be,
In △ ADM and △ BEM,
AD＝BE
∠A＝∠EBM＝45°
AM＝BM ，
So △ ADM ≌ △ BEM (SAS),
So DM = em,
So △ DEM is an isosceles triangle
As shown in the figure, in △ ABC, ∠ B = 90 °, ab = BC, BD = CE, M is the midpoint of AC edge, Verification: △ DEM is isosceles triangle
Proof: connect BM,
Because AB = BC, am = MC,
So BM ⊥ AC, and ∠ ABM = ∠ CBM = 1
2∠ABC=45°，
Because AB = BC,
So ∠ a = ∠ C = 180 ° − ABC
2=45°，
So ∠ a = ∠ ABM, so am = BM,
Because BD = CE, ab = BC, so ab-bd = bc-ce, that is, ad = be,
In △ ADM and △ BEM,
AD＝BE
∠A＝∠EBM＝45°
AM＝BM ，
So △ ADM ≌ △ BEM (SAS),
So DM = em,
So △ DEM is an isosceles triangle
As shown in the figure, △ ABC, D is a point on BC. If AB = 10, BD = 6, ad = 8, AC = 17, calculate the area of △ ABC
0
0
First of all, extend the center line AD by one time and mark it as AE, because ad = De, BD = DC, angle BDA = angle EDC, so triangle BDA and triangle EDC are congruent, because, triangle ABC area = triangle DBA + triangle ADC, triangle BDA is equal to triangle EDC, so triangle ABC area = triangle EDC + triangle ADC = triangle AEC three
In the triangle ABC, ad is the center line, ab = 17, BC = 16, ad = 15. Find the length of AC
BD=1/2*BC=8
Because AB = 17 ad = 15 BD = 8
So △ abd is a right triangle, ∠ ADB = 90 degrees
So ad is the vertical line of BC
So AB = AC = 17
In the triangle ABC, if AB = 17, BC = 16, and the center line ad = 15 on BC side, then AC=----
BD=BC/2=8
BD^2+AD^2=15^2+8^2=289=AB^2
So ad is also a vertical line, AC = AB = 17
As shown in the figure, given the triangle ABC, ab = 10, BC = 21, AC = 17, find the height on BC side
Let the height be ad, BD = X
be
10^2-X^2=17^2-(21-X)^2
(21-X)^2-X^2=17^2-10^2
(21-X+X)(21-X-X)=189
21*(21-2X)=189
42X=441-189=252
X=6
In the triangle ABC, AB is equal to 10, AC is equal to 17, and BC is equal to 21
^What do you mean?
In the triangle ABC, ab = 10, AC = 21, BC = 17 It's an obtuse triangle clockwise ABC. It's just hard to calculate. I'm equal to 8. Don't you know? I'm dizzy`` How can I know if I'm right or wrong? Maybe you're talking nonsense`
The area s = √ [p * (P-A) * (P-B) * (P-C)] can be obtained by using Helen's formula, where p = (a + B + C) / 2, a, B, C are the three sides of a triangle
Obviously, s = 84,
And S = the height on the edge of AC * AC / 2 = the height on the edge of 21 * AC / 2,
So the height on the edge of AC = 8,
Your answer is correct!
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