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Will you please help me out in solving a question as There is a right angle isosceles triangle ABC 90 degree angle at B and AC is hypotenuse. There are two points D, E in between AC such that AD:DE:EC=3:5:4 then prove that angle DBE=45degree

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestIf we take $m \equiv 0 \pmod4$, we can take $x=8$ and y as the required multiple of 4. Note that 16 has to written as 10+6.

If we take $m \equiv 1 \pmod4$, we can take $x=9$ and y as the required multiple of 4.

If we take $m \equiv 2 \pmod4$, we can take $x=10$ and y as the required multiple of 4.

If we take $m \equiv 3 \pmod4$, we can take $x=15$ and y as the required multiple of 4.

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Great!

Depending on the intent of the question, you might be missing one small case.

Note: Most people do not consider 0 a composite number.

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Oh right! 15 can be written as $6+9$.

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Will you please help me out in solving a question as There is a right angle isosceles triangle ABC 90 degree angle at B and AC is hypotenuse. There are two points D, E in between AC such that AD:DE:EC=3:5:4 then prove that angle DBE=45degree

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$m>11 \\ m = 11 + a ; ~ a \in Q\\ a = \pm n, ~ n \in Q \\ m = x + y, ~x = 11, y = \pm n$

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Can you explain what you are trying to do? I have several concerns:

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Ahh, is this sum that tough? I am doomed.

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It's actually a pretty easy problem, just give it a try. What have you tried? Where did you get stuck?

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