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 Distance to Diagonal (Posted on 2013-05-14)

Let ABCD be a parallelogram with ∠BAC = 45° and ∠ABD = 30°.

What is the distance from B to diagonal AC in terms of the
lengths of sides AB and AD?

 See The Solution Submitted by Bractals Rating: 3.0000 (1 votes)

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 re: Typo in problem statement | Comment 4 of 6 |
(In reply to Typo in problem statement by Bractals)

It's still pretty easy.
I'll call the distance sought h.
Area of triangle ABC = .5*AB*AD*sin(45º)
also
Area of triangle ABC = .5*AC*h

So .5*AB*AD*sin(45º) = .5*AC*h

By law of cosines
AC²=AB² + BC² - 2*AB*BC*cos(135º)

If you substitute this for AC as well as AD for BC the final expression for h is rather messy:

(Note: h can probably be written in terms of just AB, since the triangle ABD is uniquely determined by the ASA congruence theorem.)

 Posted by Jer on 2013-05-16 00:38:14

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