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Find the measure of the angles of triangle BDE, where BD is the altitude drawn to the hypotenuse of isosceles right triangle ABC and DE is a perpendicular bisector of BC.

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Okay, here are the steps:

* Triangle ABC is an isosceles right triangle. This means angles A and C are equal.

* BD is the altitude drawn to the hypotenuse BC. This means BDE forms another right triangle with right angle at B.

* DE is the perpendicular bisector of BC. This means DE splits BC into two equal segments.

* Since BC is split into two equal segments, angles BDE and DBE must also be equal. We can call them both x.

* In the big triangle BDE:

- Angle at B is 90 degrees (right angle)

- One other angle is x (as discussed)

- Interior angles of a triangle sum to 180 degrees

So: 90 + x + x = 180

=> 2x = 90

=> x = 45

Therefore, the measures of the angles of triangle BDE are:

BDE = 45 degrees

DBE = 45 degrees

Angle at B (90 degrees)

Does this make sense? Let me know if you have any other questions!

User Kliment Merzlyakov
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