Angle bisector theorem proof | Special properties and parts of triangles | Geometry | Khan Academy

What the angle bisector theorem is and its proof

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Wow. Great. I wasn’t coming up with that one though I tried and tried. And I had to watch the video to the end and restart it to finally get the proof. Now of course it looks almost trivial once I construct and consider that under-triangle which is similar. Thanks!!

[geraldvillanueva]

thank you!!!! i have more idea for my report!!!its so very helpful for me THANK'S

[nirvairsingh]

Have to admit it your way of teaching is simple and cool . I wish u could be my maths teacher keep up the good work man🔥🔥🔥🔥

[kaushikdr]

Basically, can you say that since
1) BCA ~ FDC (same angles) and BDC = FDC (SAS), that BCA ~ BDC?

[Its_my_plan]

sir i didn"t understand that part when you said AD = CD . The question is, is it necessary if you bisect an angle then the segment opposite to the angle will split into 2 equal part?

[holyshit]

This theorem can be used to derive u sub for R(x, sqrt(ax^2+bx+c)) integrals after completing the square

[yeshwantghotkar]

Do for external bisector of an angle of a triangle plz

[zack_]

This triangle used is almost isosceles . If cut line BC very short, say <1/3 of the current length, do these properties still hold?

[paulwang]

in fact you can use the ratio of areas of triangle ABD and CBD. This ratio equals to both AD/CD and AB/BC. In the first case they have the same height (the distance from B to AC) and in the second case they have the same height (the distance from D to AB and BC).

[tanishasagar]

Your english so good than that of maths but maths is also nice as well

[svnvoo]

can we also set the proportion as AB/BC = AD/CD?

[soumisil]

If two straight lines are perpendicular to different straight lines respectively, then prove that the angle between the two staight lines are equal to that of the perpendiculars.can anyone help me prove this theorem?

[paolasaenz]

Khan Academy can you please help me figure out the difference between a postulate and a definition's meaning in geometry please. I have test this week and I really need the help.

[namir]

Can we produce AB and join it with the line constructed by us from C which is parallel to BD

[drevil]

isnt it impossible to manipulate the length of CF. If you have that the angle of FDC is defined and the length of DC is defined and then if Angle DCF is at point C and is parallel to line BD, doesnt that mean that the angle has to already be defined so the only way the line is equal to BC, if the shape is a rhombus or did i miss something here?

[zack_]

1:06 - Talking about this theorem the triangle used for demonstration needs to be obviously different from an equilateral one like this one used here where the side-to-side ratios ARE visually same thus no need of proving.