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ABCD is a square. C' is a point on BA and B' is a point on AD such that BB' and CC' are perpendicular. Show that BB' = CC'

I don't know where to start. Use similarities?

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    Angles ABB' and BCC' are equal. – 2017-02-03
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    You can start by drawing the figure :) – 2017-02-03

2 Answers 2

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$\alpha = \alpha ‘$ because they are both complement to $\beta$.

Then, by ASA, $\triangle B’AB \cong \triangle C’BC$.

Result follows.

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    How can you prove that the angle at C' is the same as beta? – 2017-02-03
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    @GerardL. We can prove that directly but we don't because 1) As mentioned, $\alpha = \alpha'$; 2) AB = BC; 3) $\angle A = \angle C = 90^0$. Then, the two said triangles are congruent. Consequently, $\angle C' = \beta$ by corresponding measurements. – 2017-02-04
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    Will the down-voter explain why? – 2017-02-04
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    That was before his comment. Changing vote. – 2017-02-09
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I am attaching the solution please go through it. This can be done by using co ordinate geometry that I have shown in my solution.

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    How do you know that the slopes multiplied is -1? – 2017-02-03
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    I don't know it. I obtained it by calculation by the steps given above. If the product of the slopes of two lines is $-1$ then these lines are perpendicular. – 2017-02-13