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# Geometry

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Let ABCD be a trapezoid with bases AB and CD.  Let P be a point on side CD, and let X, Y be the feet of the altitudes from P to AD, BC respectively.  Prove that if AD = 5, BC = 7, AB = 6, CD = 12, and CP/PD = 1, then PX = 12*sqrt(6)/5 and PY = 12*sqrt(6)/7.

Jul 23, 2024

### 1+0 Answers

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Draw AM , BN perp to CD

Let DN = x

CN = 12 - ( AB + DN)  =  12 - (6 + x)  =  6-x

AM = BN

So....By the Pythagorean Theorem

AD^2 - DM^2  = BC^2 - CN^2

5^2 - x^2  = 7^2 - ( 6-x)^2

25 - x^2 = 49 - x^2 + 12x - 36

25 =  12x + 13

12 = 12x

x =1 = DM

Note that we have two right triangles ADM  and PDX

Angles AMD and PXD are equal = 90°

And angle ADM = angle PDX

So  triangle ADM is similar to triangle PDX

PD / AD = 6/5

So

DX / DM = 6/5

DX = (6/5)DM = (6/5) (1) = 6/5

So

sqrt [DP^2 - DX^2] = PX

sqrt [ 6^2 - (6/5)^2 ] =  sqrt [ 36*25 - 36 ] / 5 = sqrt [ 900 - 36] / 5  = sqrt [ 864] / 5  =

sqrt [ 144 * 6 ] / 5  =  12sqrt (6) / 5 = PX

Similarly

Right Triangles BCN and PCY are similar

PC/BC  = 6/7

So

CY / CN  = 6/7

CY / 5 = 6/7

CY = 30/7

So

sqrt [ PC^2 - CY^2 ] = PY

sqrt [ 6^2 - (30/7)^2] = sqrt [ 36 * 49 - 900 ] / 7  =  sqrt [ 1764 - 900] / 7 = sqrt [ 864]/7  =

sqrt [ 144 *6 ] / 7 = 12sqrt(6) / 7 = PY

Jul 24, 2024
edited by CPhill  Jul 24, 2024