Elementary 1 math triangle problem, 1st math triangle

∠f=360°-∠fga-∠fha-∠gah=360°-（180°-∠d-∠deg）-（180°-∠b-∠hcb）-(d+∠deh)=∠d+∠deg+∠b+∠hcb-∠d-∠deh=∠b-∠deg+∠hcb

The same can be said: f= d- bch+ ged

2∠f=∠b+∠d-∠deg+∠hcb-∠bch+∠ged=∠b+∠d

f=（∠b+∠d）/2

f=（∠b+∠d）/2

So when f=90°, i.e., b+ d=180°, i.e., b=180°- d= bac+ dea, ef and fc are perpendicular.

So when B> BAC, it is possible that EF and FC are perpendicular.

b: d: f=(2 b):(2 d):(b+ d), and b: d: f=2:x:3

2∠b）：（b+∠d）=2：3

b：∠d=1：2

So x=4

Because, the angle FHB = the angle F + the angle BEF

Angular FCB = Angular FCD

So, angle b = angle fhb - angle fcb = angle f + angle bef - angle fcb = angle f + angle bef - angle fcd

In the same way, angle d = angle f + angle fcd - angle bef

So, angle b + angle d = angle f + angle bef - angle fcd + angle f + angle fcd - angle bef = 2 * angle f

When De BC, EF and FC are perpendicular.

2+x=2*3

x = 4, i.e. angle d = 2 angle b.

Because b is 28°, bad=90°- b=62° and dae 16°

So bae= bad- dae=62°-16°=46° because ae bisects bac, so bac=2 bae=46°*2=92° according to the sum of the three inner angles of the triangle is 180°, get:

c=180°-∠b-∠bac=180°-28°-92°=60°

There is a rule in this question: dae=2 1( b- c) Pay attention to which is bigger and who is first between b and c.

Next, just put the numbers in.

First make an auxiliary line to connect HP, because AE bisects the angle CA, so the angle CAE = angle EAD, and the angle ACB = angle APE = 90 degrees, and AE is common, so the triangle CAE is all equal to the triangle PAE. So ce=ep. And because in the triangle CAE and the triangle AHD, there are already two equal angles, so the angle AHD = angle CHE = angle CEH, so CH = CE = EP.

Because Cd is parallel to EP, it is a diamond Ceph, so HP is parallel to CB, and HF is parallel to Pb, so HfBP is parallelogram, so HP=FB=CE (passed from above), so CE=BF. Maybe not the easiest way, drawing is the most important! But I hope you can understand and help you!

Because the angle b=45' the angle c=75' so the angle bac=60' because ae is bisector so ae bisects the angle bac so the angle cae=30'

And because the angle c=75', the angle AEC=75', because BC is high, so the angle ade=90'

So in a triangle, the angle ade = 90 degrees, the angle AEC (angle AED) = 75 degrees, so the angle dae = 15 degrees

Sadly, the first year of junior high school math stumped me who had just finished the college entrance examination

Set to n-side.

Then: The sum of the internal angles is (n-2)*180 degrees.

Since the internal angles are equally different, the internal angles=n2 multiplied by (100+140)2 are equal n=6

1. You forgot the picture.

If you make up for the brain, D is on the edge of AB, otherwise there are too many unknowns BCD plus ACD plus angle A and angle B is 180 degrees, where angle A plus angle BCD is 90 degrees, and angle BCD is 18 degrees, then angle B is 72 degrees.

If D is inside ABC, then there is a functional relationship between angle B and angle BAD with angle B + angle BAD = 72 degrees, and angle B is uncertain).

2. The distance from point A to BC is the height of the edge of BC.

In this way, there is a formula for the area of the base multiplied by the height of two.

Then the high is 26*2 10=

1) Area of ABEs = 6x5x1 2=15 square centimeters Area of ABC = 2x6x5x1 2=30 square centimeters2) S shade = 1 4 x S ABC=1 square centimeters3) Length of AC=5-2=3 cm.

2.If two lines are perpendicular to the same line, then the two lines are parallel.

2=65°5 questions are problematic.

No time. I'll write the rest when I have time.。。