drawing a triangle with specifc arcs

I am about to write down some calculations in LaTeX. To make them more easily understandable, I want to add a geometric figure. Unfortunately, I am really new to Tikz. While I am able to draw lines, rectangles, triangles, name points, etc. I haven't yet drawn something with a precise angle.

I want to draw a triangle with the vertexes A, B and C and a angle bisector (dividing the angle at B).

I know that the angle at the vertex A = 44°, B = C = 68° and that the distances AB = AC.

\begin{tikzpicture}
\begin{scope}
\tkzDefPoint(0,0){B};
\tkzDefPoint(1,-3){C};
\tkzDefPoint(-1,-3){A};

\clip (-2,0.5) rectangle (2,-4);

\draw[step=.5cm,gray,very thin] (-4,-4) grid (4,4);
\draw (-4,0) -- (4,0);
\draw (0,-4) -- (0,4);

\draw[thick] (A) -- (B) -- (C) -- cycle;
\tkzLabelPoints[above](B);
\tkzLabelPoints[below](C);
\tkzLabelPoints[below](A);
\draw (B) circle [radius = 3.2pt];

\end{scope}
\end{tikzpicture}


But how do I specify that I want the base angle at the Vertex B to be 68° exactly?

Update: In the manual I have found the following method to draw angles: (45:1cm). Unfortunately, this command uses as a start point whatever coordinate have used last. But how do you specify, that I want it to start from B?

• You can use the + and ++ Syntax to specify relative coordinates: (0,0) -- +(1,0) -- ++(0,1) means from origin go 1cm right and 0cm up, than return reference position to last absolute coordinate (in this case (0,0)),, then go 1cm up and 0cm left and fix this position as new reference (that's the ++ notation). If I'm not clear enough, look it up in the Manual, it's well explained. :) – Guilherme Zanotelli Dec 14 '16 at 15:45

Two methods provided (code is self explanatory - I think):

\documentclass[tikz]{standalone}
\usetikzlibrary{angles,quotes}
\usetikzlibrary{shapes.geometric}
\begin{document}
\begin{tikzpicture}
% Draw it manually using polar coordinates (no libraries needed, I used quotes and angles just to draw the angles):
\draw (0,0) coordinate (B) -- ++(68:3cm) coordinate (A) -- ++(68+44:-3cm) coordinate (C) -- cycle
pic["$B$", draw] {angle=C--B--A}
pic["$A$", draw] {angle=B--A--C}
pic["$C$", draw] {angle=A--C--B};

%Using the shapes.geometric library with the "isosceles triangle" shape:
\node[isosceles triangle,
isosceles triangle apex angle=44,
minimum width={6cm*cos(68)},
draw, anchor=left corner, rotate=90] at (3.2,0) {};
\end{tikzpicture}
\end{document}


• A question to "isosceles triangle" shape: How to you get the coordinates of the remaining two corners so that I am able to label them and stuff? – Willi Dec 14 '16 at 15:48
• There are, besides from all border anchors (e.g. A.130) and text node anchors, the A.apex, A.left corner and A.right corner - assuming A to be the name of the node. You can get even more Information from more available shapes inside the manual in the Libraries Part, section Shapes. – Guilherme Zanotelli Dec 14 '16 at 15:52
• Thank you very much. I have chosen the manual version for the reason of practicing tikz. Unfortunatly I cannot use pic["$B$", draw] {angle=C--B--A} pic["$A$", draw] {angle=B--A--C} pic["$C$", draw] {angle=A--C--B} as they will cause a bunch of error. Sadly, I liked that notation for adding \alpha = 68° instead of the vertex name. – Willi Dec 14 '16 at 16:02
• @Willi, do note that to use such Notation you must do some things: (1) load libraries quotes and angles, (2) You have to name the three coordinates, (3) if you're using pic outside the scope of a path (stuff between \draw and ;) you have to use \pic instead (and place the semicolon at the end) pic is like node, if it's in the middle of the path you use node if it's by itself, you use \node. If the errors persist, open a new question, that's not normal behavior. :) – Guilherme Zanotelli Dec 14 '16 at 16:49
\documentclass[border=2mm,tikz]{standalone}
\usetikzlibrary{calc}
\begin{document}

\begin{tikzpicture}
\draw[step=.5cm,gray,very thin] (-3,1) grid (2,-3);
\draw[thick](0cm,0cm)coordinate[label={B}](B)
--(224:3cm)coordinate[label =left:A](A)
--++(3cm,0)coordinate[label=right:C](C)
--(B)--cycle (A)--($(B)!.5!(C)$);
\draw (-3,0) -- (2,0);
\draw (A) circle [radius = 2pt];
\clip (-.5,.5) rectangle (2,-2.5);