# ABCD is a square. The diagonals AC & BD cut at A. From B a perpendicular is drawn to the bisector /_DCA and it cuts AC at P and DC at Q. Can you prove that DQ = 2PE?

##### 1 Answer
Oct 24, 2017

Yes, I think I can... #### Explanation:

...imperfectly constructed diagram above, but hopefully it communicates the idea.

If I understand it correctly, points B, P, and E form a right triangle, where points BEP define the right angle.

We can draw another line through point D, perpendicular to diagonal BD. This new line is therefore necessarily parallel to diagonal AC.

Now extend your line from B, through P, and Q, and extend that until it intersects the line we drew in the step immediately above. I'll call the point where it does that Z

They share a common angle - angle EBP is the same as angle DBZ.

So, this new triangle is similar to original right triangle EBP.

Side BD is twice the length of BE, so therefore every side of our new triangle DBZ is twice the length of the corresponding side on EBP.

Therefore DZ is twice the length of EP. (DZ = 2PE).

If we can show that segment DZ = DQ, then we will have proved the proposition.

This will be true if triangle DQZ is isosceles, where angle DQZ = DZQ.

Note the way in which line BP intersects diagonal AC
Angles BPE and CPQ, being the opposite angles defined by the intersection of two lines, are therefore equal.

Now examine the point where line BP intersects the bisector of angle ACD. (I don't think we've labeled it yet, call it R).

Since this line bisects angle ACD, and since we've specified that line BP (also line BQ) intersects the bisector at a right angle, we have two similar right triangles PCR and RQC..

Since they are similar right triangles, angle RPC is equal to RQC.

Angles DQZ and RQC, are opposite angles formed by the intersection of two lines, and are therefore equal.

Angle DQZ is therefore equal to angle EPB, and, since they are similar triangles, angle DZB is also equal to angle EPB.

So, since DQZ = DZB, triangle DQZ is isosceles, and line segment DQ equals DZ, which we have already shown to be of length 2EP.

Therefore, segment DQ also equals 2EP.