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

In this page, a short and to-the-point overview of constructing triangles with angle-side-negative 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 `/_\ ABC` with given "side-angle-difference between other two sides" `bar(AB)=4`cm, `/_A=50^@`, and `bar(AC)-bar(BC) = -1`cm OR `bar(BC)-bar(AC) = 1`cm. • Construct a line and mark `bar(AB)` measuring `4`cm

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

• measure `bar(BC)-bar(AC)` `=1`cm and draw an arc cutting the ray at `D` at the back-end of the ray, as shown in the figure.

This is directly constructed from the given parameters.

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

The point `C` is visualized in `vec(AD)` such that `bar(BC)-bar(AC)=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(BC)-bar(AC)` at the given angle `/_A`. The point `C` is visualized in `vec(AD)` such that `bar(BC)-bar(AC)=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 the figure to visualize. Which of the following property helps 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(BC)-bar(AC)` at the given angle `/_A`. The point `C` is visualized in `vec(AD)` such that `bar(BC)-bar(AC)=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 to be 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` can be 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(BC)-bar(AC)` at the given angle `/_A`. The point `C` is visualized in `vec(AD)` such that `bar(BC)-bar(AC)=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 to be 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-Negative 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*