Question 386789: Prove that if in the triangle ABC with right angle at A, there is a point D on CB so that AC equals AD equals DB, then angleB equals 30 degrees.
This has completely stumped me.
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! I don't know how to insert my own pictures into this solution, so you'll have to bear with me and try to draw the triangle and its angles:
Let angle ABD = BAD = x (since triangle BAD is isosceles). Since angle ABD + angle BAD = angle CDA (since the sum of 2 angles in a triangle is equal to the exterior angle on the third vertex), we can let angle CDA = 2x. Also, since triangle CAD is isosceles, angle ACD is also equal to 2x.
From the diagram, angles ACD and ABD are two of the angles in triangle ABC which add up to 90. Since ACD = 2x and ABD = x, then 2x + x = 90 --> x = angle ABD = 30 degrees.
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