In the triangle ABC, ad is perpendicular to D, angle B = 2, angle C. It is proved that ab + BD = CD

Proof: take a point E on DC, make BD = ed, connect AE
Because angle AED = angle c + angle CAE, angle CAE = angle c = half angle B, so AE = EC = ab. so AB + BD = ed + CE = CD

In the known triangle ABC, the angle c = 2, the angle B, BC = 2Ac, and the proved angle a = 90 degrees

Take the midpoint D of BC, connect ad, the triangle CAD and DAB are isosceles triangles, so the two diagonal angles are equal, so the angle a = angle B + angle c is equal to 90 degrees

As shown in the figure, in the known triangle ABC, angle a = 2, angle B, CD is the bisector of angle ACB Verification: BC = AC + ad

Proof: extend CA to e, make AE = ad, connect ed
∵AE=AD,∴∠E=∠ADE,
∴∠CAD=∠E+∠ADE=2∠E,
∵∠CAD=∠2∠B
∴∠E=∠B,∠ECD=∠BCD,AD=AD
∴△ECD≌△BCD
∴BC=EC=AC+AE=AC+AD

In the triangle ABC, the angle ACB = 2 angle B, BC = 2Ac, it is proved that the angle a is equal to 90 degrees

Take the midpoint of BC, connect a and the midpoint, and you can prove it. They are all expressed in multiples of angle B

As shown in the figure, in the triangle ABC, the angle ACB = 2 angle ABC. It is proved that the square of AB = the square of AC + AC × BC Can't send the picture, the picture is a triangle ABC, three lines, will speed answer, Well, the first floor, can we use similar to do, your practice we have not learned

You can do it with similarity: you can look at my picture while you look at it
Make an angular bisector through point C and intersect AB at point D,
Then △ CDA ∽ BCA
So AB / AC = BC / CD
Because ∠ B = ∠ BCD, BD = CD
So AB / AC = BC / BD
So AB * BD = AC * BC
Because BD = ba-da
So AB * (ba-da) = AC * BC
AB²=AB*DA+AC*BC
And △ CDA ∽ BCA
So DA / Ca = Ca / ab
So AB * Da = Ca
So AB 2 = Ca 2 + AC * BC
I wonder if there is a simpler one. This is a way

In △ ABC, ad is the bisector of angle, ad = BD, ad = BD, ab = 2Ac. It is proved that △ ACB is a right triangle D for de ⊥ AC ∵ ad is the angular bisector ∴BD/CD=AB/AC ∵AB=2AC ∴BD=2CD Let CD = a, then BD = 2A ∵AD=BD ∴AD=2a ∵DE⊥AC ∴∠AED=90°,∠DAE=30°=∠BAD=∠B Ψ C = 90 °, that is, C and D coincide The △ ACB is a right triangle BD/CD=AB/AC

1. Theorem: if ad is a bisector of △ ABC ∠ a, then
The following is a brief proof: extend Ba to C ', make AC' = AC, connect CC ', from the two base angles of the isosceles triangle ACC' are equal, equal to ∠ A / 2, we can get CC '∥ Da, △ bad

It is known that in △ ABC, ab = 2Ac, ad is the angular bisector of ∠ BAC, and ad = BD

In the △ ADC and △ ade, ad = ad ∵ ad = BD,  AE = be = 12ab, ∵ ad = BD, ∵ AB = 2Ac,  AE = AC, ∵ ad bisection  BAC, ∵ CAD ∵ CAD = ed, in △ ADC and △ ade, ad = ad ⊙ ade (SAS), \\\\\\\\\\\\it's a good idea

It is known that in △ ABC, ab = 2Ac, ad is the angular bisector of ∠ BAC, and ad = BD

In the known triangle ABC, BC = 2Ac, ad is the midline, AE is the midline of ABD Confirmation: AC = 2ae Wrong BC=2AB To use similarity?

It is proved that: ∵ BC = 2Ab, D is the midpoint of BC, e is the midpoint of BD  be / AB = AB / BC = 1 / 2 ? B = ∵ B ∵ B ∵ B ∵ AE / AC = AB / BC = 1 / 2 ∵ AC = 2ae

Triangle ABC, the angle B is equal to 2 times the angle c, BC is equal to 2Ac?

∵BC-AC

In the triangle ABC, ad is perpendicular to D, angle B = 2, angle C. It is proved that ab + BD = CD