Ad is perpendicular to BC, CE is perpendicular to AB, ad is 8, CE is 7, the sum of AB and BC is 45, the area of triangle ABC

Ad is perpendicular to BC, CE is perpendicular to AB, ad is 8, CE is 7, the sum of AB and BC is 45, the area of triangle ABC

1 / 2 ad * BC = 1 / 2 CE * ab
AB：BC=AD：CE=8：7
AB=8/(8+7）*45=24
S=1／2*AB*CE
=1/2 *24*7
=84

In the triangle ABC, ad is perpendicular to BC, CE is perpendicular to AB, ad = 8cm, CE = 7cm, AB + BC = 21cm. What is the area of triangular ABC? Please help the netizens, thank you!

S=1/2*bc*ad=1/2*ab*ce
∴bc/ab=ce/ad=7/8
∵ab+bc=21
∴ab=56/5
bc=49/5
∴S=1/2*49/5 * 8=196/5

It is known that we can find a point P in the plane of the equilateral triangle ABC so that △ ABP △ ACP △ CBP are isosceles triangles. How many such p points are there? It's better to have a picture. The answer is 10, but I don't know which six are left besides the one in the triangle and the four with equilateral triangles on each side,

Each vertical bisector has three intersections, plus the vertical center of the triangle, a total of 10

Given that the triangle ABC, AB is greater than AC, ad is the high on the edge of BC. It is proved that ab squared minus AC squared is the difference between BC times BD minus DC

In the RT triangle ABC,
ab2=bd2+ad2
In RT triangle ADC
ac2=ad2+cd2
ab2-ac2=bd2+ad2-(ad2+cd2)
=bd2-cd2
=（bd+cd)（bd-cd)
=bc(bd-dc)

As shown in the figure, △ ABC, ab = BC = AC, ∠ B = ∠ C = 60 °, BD = CE, where ad and be intersect at point P, then the degree of ∠ ape is () A. 45° B. 55° C. 75° D. 60°

In △ abd and △ BCE,
AB＝BC
∠ABD＝∠BCE
BD＝CE ，
∴△ABD≌△BCE（SAS），
∴∠BAD=∠CBE，
∵∠APE=∠ABE+∠BAD，∠ABE+∠CBE=60°，
∴∠APE=∠ABC=60°．
Therefore, D

Ad is the center line of △ ABC, and E is the midpoint of AD. if ∠ DAC = B, BD = CE, try to explain △ ace ∽ bad

Because ad is the center line of △ ABC,
So BD = DC
Because BD = CE
So DC = CE
So angle ADC = angle CEA
So angle ADB = angle AEC
Because ∠ DAC = ∠ B BD = CE
So △ ace ∽ bad

In the triangle ABC, it is known that ad is the bisector of the angle and AE is the high. If the angle B = 42 degrees and the angle c = 66 degrees are known, the degree of the angle DAE is calculated

∠BAC=180-∠B-∠C
           =180-42-66=72
∠BAE=90-∠B=90-42=48
∠BAD=∠BAC/2=72/2=36
∠DAE=∠BAE-∠BAD=48-36=12

It is known that in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at point D, be bisection ∠ ABC, intersection ad at point m, an bisection ∠ DAC, intersection BC at point n Verification: the quadrilateral amne is rhombic

Proof: ∵ ad ⊥ BC,
∴∠BDA=90°，
∵∠BAC=90°，
∴∠ABC+∠C=90°，∠ABC+∠BAD=90°，
∴∠BAD=∠C，
∵ an bisection ∠ DAC,
∴∠CAN=∠DAN，
∵∠BAN=∠BAD+∠DAN，∠BNA=∠C+∠CAN，
∴∠BAN=∠BNA，
∵ be bisection ∵ ABC,
∴BE⊥AN，OA=ON，
Similarly, OM = OE,
The quadrilateral amne is a parallelogram,
The parallelogram amne is a diamond

In triangle ABC, Math homework help users 2016-11-29 report Use this app to check the operation efficiently and accurately!

Angle AEM = 90 degrees angle Abe = 90 degrees angle CBE = angle BMD = angle Ame
Am = AE, and angle man = angle can, so: an is the vertical line of me
And an is perpendicular to be, and angle Abe = angle CBE, so: me is the perpendicular line of an
In other words, an and me are equally divided vertically
So the quadrilateral amne is a diamond

As shown in the figure, ad is the height on the hypotenuse BC of the R triangle ABC,

First of all: ∠ B = ∠ DAC, because the sum of the two angles and ∠ C is 90 degrees, and from an and be are angle bisectors, we can know that: ∠ Abe = ∠ NAE, so the triangle OAE is a right triangle, ∠ OAE is 90 degrees, that is, Ao is not only the angular bisector of the triangle ame, but also its vertical line, so am = AE, Mo = OE