The puzzle of the bisector of the first two corners, master

Do dp perpendicular ab, dq perpendicular ac, because it is a bisector, then dq is equal to dp, because there is a ninety degrees, and because de is equal to df, so the triangle edp congruent triangle feq. So the angle deb is equal to the angle dfa, so: aed= dfc

The perpendicular line of the D to do AB crosses AB to M

The perpendicular line of the D to do AC crosses the AC to N

Vertical dm = perpendicular dn

And the triangle dem and triangle dfn are both right triangles because in right triangles.

de=df,dn=dm

So the right triangle dem is equal to the right triangle dfn (hl theorem, as we called it at the time).

So med=nfd

This is AED= DFC

By making dg perpendicular to ae to g, and dd perpendicular to ac to h can be proved: adg is equal to adh

So: dg dh

In RT DGE vs. RT DHF.

de＝df dg=dh

The Pythagorean theorem yields ge=hf

So rt dge is exactly equal to rt dhf

So aed= DFC

Take a little g on the ac of bac so that ag=ae is used as an auxiliary line dg according to the meaning of the title de=dg aed= agd i.e. = fgd in dfg because de=df so df=dgdfg is an isosceles triangle fgd=== dfg i.e. = dfc so: aed= dfc

Proof: The point d is dm perpendicular ab to m, and the point d is dn perpendicular ac to n, because ad bisects bac dm vertical ab dn vertical ac, so dm = dn dme = dnf

Because de=df

So the triangle dme is equal to the triangle dfn

So aed= DFC

It is known that the angular bisector AD of BAC is known, and because DE=DF, the point E is on AB and the point F is on AC, so DE is perpendicular AB, and DF is perpendicular AF.

So aed= DFC

Well, is this topic wrong, if it is a request for a certificate ad deuce caf I can make it, regardless of your topic, first attach my results :

As shown in the figure, first make three perpendicular lines, the intersection points are g, h, i respectively, and then according to the axiom or theorem of the angle bisector, we can get dg=dh, dg=di, so, di=dh, and ad is the common hypotenuse of the right triangle adh and the right triangle adi, so these two triangles are congruent, and then fad= dac, so ad is the angle bisector of caf.

Can I ask if the picture is in**? How to do a geometry problem without a diagram?

You sent the picture to the night. Eldest brother.

Bisector of de BC, ABC, ACB.

adc=∠bcd=∠acd,ad=ac=6,∠aeb=∠cbe=∠abe,ae=ab=√(10²-6²)=8,de=ad+ae=6+8=14

sabc=8*6/2=24

Let the height on the BC edge be H, the height on the DE edge of DFE be H1, and the height on the BC edge of BFC be H2, then H1+H2=H.

h=24*2/10=

de bc, edf= fcb, def= fbc, dfe bfc,bc de=h2 h1 (the ratio of the corresponding sides of a similar triangle must be equal to the ratio of the height of the corresponding edges).

10 14 = (, h1 = , def area = 14*

Angle 1, Angle 2, Angle 3, Angle 4

Flat angles and 180

Triangle inner angle and 180

The sum of the inner angles of the trilateral type is 180 degrees, and the sum of the internal angles of the quadrilateral type is 360 degrees, and if the angles are equal to the angles, it should be possible to prove that + 180 degrees.

By proving the bisector theorem, the inaccuracy of your drawing, and then there's the irregularity of the letters that represent the angles, and the angles can be either a lowercase letter or a three-case letter, or a number, or a Greek letter

The paracentric theorem is used to prove that the perpendicular lines from E to three sides are made, and EM is perpendicular to BCD

EN Vertical AC

el vertical baf

From the properties of the angular bisector, em=en, em=el, en=el

Yes, AE is the angular bisector of FAC.

The symbols are given according to your title, but I suggest that you use uppercase letters to indicate points, and lowercase letters are generally used to indicate the length of the line segment.

AF divides EAD equally

The perpendicular distance from any point on the bisector of the DFH angle to both sides of this angle is equal in ADF and AHF.

daf＝∠haf

adf＝∠ahf

df hf adf congruent ahf

Ad AH is in ABE and AHG.

eab＝∠gah

ab＝ahabe＝∠ahg

ABE congruent AHG

ae＝agbe＝gh

df＝hfbe＋df＝gh＋hf

GF requires BE+DF=AE

Ag GF is also required

AGF is an equilateral triangle.

ag＝gfbe+df=ae

To make up the triangle ADF to the side of AB, that is, AD coincides with AB, you only need to prove that the new triangle ABF is an isosceles triangle, and the angle Eaf' is the same as the angle AED, and you know how to prove it.