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Think about the triangle's perimeter or area. How can you involve x in these concepts?
x=AB+AC-BC/2 or x=AB* AC/AB+AC+BC
To find an expression for x, we can use the fact that the area of â–ł ABC is equal to the area of the three smaller triangles, â–ł ABD, â–ł ADC, and â–ł BDC that's inside of â–ł ABC.
Area=1/2bh Since â–ł ABC is a right triangle, the base and height will be the length of its legs, AC and AB. For â–ł ABD, â–ł ADC, and â–ł BDC, the height is the perpendicular segment from the incenter. According to the Incenter Theorem, the incenter of a triangle is equidistant from the sides of the triangle. Therefore, the height of all of these triangles are x.
Now we can determine expressions for each triangle's area.
Triangle | b | h | 1/2bh | Area |
---|---|---|---|---|
â–ł ABC | AB | AC | 1/2(AB)(AC) | AB* AC/2 |
â–ł ABD | AB | x | 1/2(AB)(x) | AB* x/2 |
â–ł ADC | AC | x | 1/2(AC)(x) | AC* x/2 |
â–ł BDC | BC | x | 1/2(BC)(x) | BC* x/2 |
Substitute expressions
LHS * 2=RHS* 2
Factor out x
.LHS /(AB+AC+BC).=.RHS /(AB+AC+BC).
Rearrange equation
We can also write an expression for x by using the perimeter of the triangle. We have already established that the incenter, D, is equidistant, x, to each side. From the diagram, we can also identify a square with a side of x since it's a quadrilateral with four right angles and two congruent adjacent sides.
Also, since D is the incenter of â–ł ABC, it means that BD and CD are angle bisectors. This gives us enough information to show congruence between two pairs of triangles using the AAS Congruence Theorem.
Let's identify the remaining corresponding sides in each pair of congruent triangles and label them.