What is the measure of the acute angles of an isosceles right triangle?
See solution.
Practice makes perfect
We wish to prove that the two trinagles are similar. Recall that an isosceles right triangle has a 90^(∘) angle and two acute base angles that are congruent. If we call the measures of these angles m∠ a, we can write the following equation.
m∠ a+m∠ a=90^(∘)
Let's solve this equation.
Since they are both right triangles, both triangles have a right angle. We have just shown that, since each triangle is a right isosceles triangle, each triangle also has two 45^(∘) angles. Therefore, the angles from one triangle match the angles from the second triangle. This makes the triangles the same shape, and, by definition, similar.
Alternative Solution
Prove by mapping
We can also prove that the figures are similar by mapping one triangle onto the other. The triangles are facing the opposite way. Therefore, if we reflect △ ABC in AB, they will obtain the same orientation.
Next, we will translate △ ABC' so that the right angles map onto each other. A translation is a rigid motion which means it preserves length and angle measures. Therefore, AC' will map onto RT and AB onto RS.
From the diagram, we see that the length of A'B' is j and the length of the corresponding side, RS, is k. Dividing these lengths, we can determine the scale factor we need when dilating:
Scale Factor: k/j
Using A' as the center of dilation, we can dilate △ A'B'C'' to the same size as △ RST if we multiply its sides with a scale factor of kj.
A'C'':& k/j* j=k [0.8em]
A'B':& k/j* j=k
Having dilated △ A'B'C'' with a scale factor of kj, we can see that △ A''B''C''' maps onto △ RST.
