__maths__>__Construction / Practical Geometry (High)__>__Construction of Triangles with Secondary Information__### Construction of Triangles using Angle, Side, difference between 2 Sides

In this page, a short and to-the-point overview of constructing triangles with angle-side-difference between two sides is provided. Difference between two sides is a secondary parameter.

The reasoning on how to approach the problem and how the procedure works are provided.

*click on the content to continue..*

To construct a triangle, "side-angle-difference between other two sides" is provided. Which of the following is *directly* constructed using the given parameters?

- the triangle `/_\ ABC`
- the given angle with two line segments of given side and difference between sides
- the given angle with two line segments of given side and difference between sides

The answer is "the given angle with two line segments of given side and difference between sides"

To construct a triangle `/_\ ABC` with given "side-angle-difference between other two sides" `bar(AB)=4`cm, `/_A=50^@`, and `bar(AC)-bar(BC) = 1`cm. • Construct a line and mark side `bar(AB)` with a compass measuring `4`cm

• measure `50^@` angle and draw a ray from point `A`

• measure `bar(AC)-bar(BC)` `=1`cm in a compass and draw an arc cutting the ray at `D`

This is directly constructed from the given parameters.

* With given "side-angle-difference between two sides" we constructed `bar(AB)`, and `bar(AD)=bar(AC)-bar(CB)` at the given angle `/_A`.*

The objective now is to locate the point `C` in the line `vec(AD)`.

Consider points `p`, `q`, `r`, `s`, and `t` on line `vec(AD)`. Which of the following property is noted for points `p` and `t` in the line `bar(AD)`?

- `bar(Ap)- bar(pB) < bar(AD)` as evident from the figure
- `bar(At)- bar(tB) > bar(AD)` as evident from the figure
- both the above
- both the above

The answer is "both the above". It is noted that at a single position in `vec(AD)`, the point `C` is located such that `bar(AC)-bar(CB)=bar(AD)`.

*With given "side-angle-difference between two sides" we constructed `bar(AB)`, and `bar(AD)=bar(AC)-bar(CB)` at the given angle `/_A`.*

The point `C` is visualized in `vec(AD)` such that `bar(AC)-bar(CB)=bar(AD)`. Which of the following is an useful observation that will help to locate the point `C` along `vec(AD)`?

- the line segments `bar(CD)` and `bar(CB)` are equal
- the triangle `/_\ BCD` is isosceles
- both the above
- both the above

The answer is "both the above".

* With given "side-angle-difference between two sides" we constructed `bar(AB)`, and `bar(AD)=bar(AC)-bar(CB)` at the given angle `/_A`. The point `C` is visualized in `vec(AD)` such that `bar(AC)-bar(CB)=bar(AD)`. It is observed that `/_\BCD` is isosceles and only angle `/_D` and base `bar(BD)` are available. *

The position of point `C` is not yet marked. It is shown in figure to visualize. Which of the following property help in identifying the point `C`?

- two angles of an isosceles triangles are equal. `/_CDB = /_DBC`
- a perpendicular bisector on the base of isosceles triangle passes through the third vertex
- both the above
- both the above

The answer is "both the above".

*With given "side-angle-difference between two sides" we constructed `bar(AB)`, and `bar(AD)=bar(AC)-bar(CB)` at the given angle `/_A`. The point `C` is visualized in `vec(AD)` such that `bar(AC)-bar(CB)=bar(AD)`. It is observed that `/_\BCD` is isosceles and only angle `/_D` and base `bar(BD)` are available. The position of point `C` is not yet marked. It is shown in figure to visualize.* Using the property that

two angles of an isosceles triangles are equal. `/_CDB = /_DBC`

The angle `/_CDB` is copied using a compass to mark line `vec(BE)`

The point of intersection of rays `vec(AD)` and `vec(BE)` is the point `C`.

`/_\ABC` is constructed.

*With given "side-angle-difference between two sides" we constructed `bar(AB)`, and `bar(AD)=bar(AC)-bar(CB)` at the given angle `/_A`. The point `C` is visualized in `bar(AD)` such that `bar(AC)-bar(CB)=bar(AD)`. It is observed that `/_\BCD` is isosceles and only angle `/_D` and base `bar(BD)` are available. The position of point `C` is not yet marked. It is shown in figure to visualize.* Using the property that

the perpendicular bisector on the base of isosceles triangle passes through the third vertex,

The line `vec(MN)` is constructed as the bisector of `bar(DB)`. The point of intersection of `vec(AD)` and `vec(MN)` is the point `C`.

`/_\ABC` is constructed.

**Construction of Side-Angle-Difference between 2 Sides of triangle** : Visualize that at point `C`, the `/_\ BCD` is isosceles triangle.

Locate point `C` with one of the following methods.

• copy the angle

• draw perpendicular bisector on the base.

*slide-show version coming soon*