It is known that in the triangle ABC, D is a point on AB, be is perpendicular to ad, CF is perpendicular to ad, and perpendicular feet are e and f respectively If ad is the central line of the triangle ABC, be = CF is proved If ABC is a triangle of ABC, prove that it is a triangle fast

It is known that in the triangle ABC, D is a point on AB, be is perpendicular to ad, CF is perpendicular to ad, and perpendicular feet are e and f respectively If ad is the central line of the triangle ABC, be = CF is proved If ABC is a triangle of ABC, prove that it is a triangle fast

It is proved that ad is the center line of the triangle ABC, and BD = DC, be, CF are perpendicular to ad respectively, so be is parallel CF, so the angle EBD = angle FCD, angle bed = angle CFD = 90 degrees, so the triangle bed is congruent with triangle CFD, so be = CF2 prove: because: angle BDE = angle CDF (equal to vertex angle), angle CFD = angle bed = right angle, so triangle bed is similar

As shown in the figure, △ ABC, ad, be and CF are three midlines, which intersect at point O. please judge the relationship between the area of △ AOF and the area of △ AOE according to the above conditions, and explain your reasons

The area of △ AOF is equal to that of △ AOE;
Reason: ∵ ad, be, CF are three midlines,
∴S△ABD=S△ADC=S△ACF=S△BCF=S△ABE=S△BCE=1
2S△ABC，
∴S△BOD=S△AOE，S△AFO=S△COD，
∵BD=CD，
∴S△BOD=S△AOE=S△AFO=S△COD，
The area of △ AOF is equal to that of △ AOE, and the same bottom and height

If △ ABC is an equilateral triangle and ad = be = CF, then the triangle DEF is A equilateral △ B isosceles △ C any △ D right angle △

A. Because ad = be = CF, and because △ ABC is an equilateral triangle, BD = EC = AF, and angle DAF = angle DBE = angle DAF, so triangle DAF is equal to triangle, bed is equal to triangle EFC, so de = DF = EF, since all three sides are equal, of course it is a!

As shown in the figure, in the triangle ABC, ad, be, CF are the bisectors of angles, and the intersection points are points g, GH ⊥ BC. Try to explain the reason why ∠ bgd = CGH

∠ bgd is the outer angle of the triangle AGB
∠BGD = 1/2 ∠A + 1/2 ∠B
∠CGH = 90 - 1/2 ∠C = 1/2(180 -∠C) = 1/2 (∠A+∠B)
therefore
∠BGD=∠CGH

As shown in the figure, in △ ABC, ad is the angular bisector of ∠ BAC. It is proved by sine theorem that AB / AC = BD / DC Look at the picture

From the sine theorem,
In triangle abd
BD/sinBAD = AB/sinADB
DC/sinCAD = AC/sinADC
Sinbad = sincad
sinADB = sinADC
One form is compared with the above two formulas
The answer is proof

Sine theorem AB / AC = BD / DC It is proved by sine theorem that AB / AC = BD / DC

If the outer angle of a is 2a, then the angle CAD = a, the angle BAC = pai-2a; AB / BD = sind / sin (pai-a); AC / DC = sind / Sina; sin (pai-a) = Sina; therefore, AB / BD = AC / DC, the original formula can be proved
A there are two ways to draw the outer corner. The reason is the same

In Δ ABC, it is proved by sine theorem that AB / AC = BD / DC

From the sine theorem, it can be concluded that: sin ∠ cab / CD = sin ∠ CDA / AC; sin ∠ bad / BD = sin ∠ ADB / AB;
Because sin ∠ ADC = sin ∠ ADB; sin ∠ CAD = sin ∠ bad; so AC / CD = AB / BD; so AB / AC = BD / CD

In △ ABC, BD is the angular bisector of ∠ ABC. It is proved by sine theorem that AB / BC = ad / DC

According to the sine theorem, in △ abd, AB / sin ∠ BDA = ad / sin ∠ abd. In △ DBC, BC / sin ∠ BDC = DC / sin ∠ DBC is divided by two formulas, and (AB / BC) * (sin ∠ BDC / sin ∠ BDA) = (AD / DC) * (sin ∠ DBC / sin ∠ ABD) because ∠ BDA and ∠ BDC are complementary angles each other, sin ∠ BDA = sin ∠ BDC, i.e. si

In the triangle ABC, the bisector ad of the outer angle of angle A and the extension line of BC intersect at point D. It is proved that BD ratio DC is equal to AB to AC

If the outer angle of a is 2a, then the angle CAD = A and the angle BAC = π - 2A;
AB / BD = sind / sin (π - a); AC / DC = sind / Sina; sin (π - a) = Sina;
∴AB/BD=AC/DC
∴BD/DC=AB/AC

As shown in the figure, in the triangle ABC, the angle B is equal to 2, the angle c is perpendicular to BC, and the perpendicular foot is d. two methods are used to prove that AB and BD are equal to DC

It is proved that: take point E on the extension line of CB, make be = AB, and connect AE
∵BE＝AB
∴∠BAE＝∠E
∴∠ABC＝∠BAE+∠E＝2∠E
∵∠ABC＝2∠C
∴∠E＝∠C
∴AE＝AC
∵AD⊥BC
/ / ed = CD (three wires in one)
∵ED＝BE+BD
∴ED＝AB+BD
∴CD＝AB+BD