Problem 28
Let I be the incenter of the a triangle ABC, where C = 60°. Let the internal angle bisectors of angles A and B meet BC and AC at D and E respectively. Prove that:
ID = IE
Here’s a simulation that illustrates the problem.
Try dragging A and C. Are ID and IE always equal? How can we prove the same?
Solution
Here are some concepts that might help you in solving the problem.
Incenter of a Triangle
The point of concurrency of the internal angle bisectors of a triangle is known as its incenter. The incenter is always equidistant from the three sides of the triangle.
Here’s a simulation that shows the incenter. Tap the play button to start the animation.
You can try dragging the vertices of the triangle. Is the incenter at the same distance from each side?
Congruent Triangles
If the corresponding sides and angles of two triangles are equal, then they are said to be congruent. And, if we’re able to prove that two triangles are congruent, then their corresponding parts (sides and angles) are equal.
There are various ways to prove the congruence of two triangles, one of them being the Right angle-Hypotenuse-Side or RHS rule.
That is, if for two triangles ABC and DEF that AB = DE, BC = EF, and ∠B = ∠E, then the two triangles would be congruent.
And, this would mean that AC = DF, ∠A = ∠D, and ∠C = ∠F (i.e., the corresponding parts are equal).
You can observe this in the following simulation.
Drag E and F such that AB = DE and AC = DF. Does BC equal EF in that case?
Problem Source: CutTheKnot
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