Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 729 details.

## Wednesday, February 15, 2012

### Problem 729: Triangle, Altitude, Orthocenter, Circumcircle, Perpendicular, Midpoint

Labels:
altitude,
circumcircle,
midpoint,
orthocenter,
perpendicular,
triangle

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http://img713.imageshack.us/img713/4057/problem729.png

ReplyDeleteLet BH cut the circle at L . ( see picture)

Let P is the projection of D over AC and Q is the point of symmetry of D over AC

We have NH= NL ( property of orthocenter)

And E, F and P are collinear ( Simson line)

HLQD is isosceles trapezoid ==> N, S , P are collinear

HSL is a isosceles triangle => ∠ (NHS)= ∠ (NLS)= ∠ (SQP) ( line HL // DQ)

Quadrilateral DFPC is cyclic => ∠ (FPD)= ∠ (FCD) ( both angles face arc FD)

But ∠ (FCD)= ∠ (BLD)= ∠ (NLS) ( angles face same arc BD)

So∠ (FPD)= ∠ (NLS)= ∠ (SQP) => line EFP // line HSQ

Since P is the midpoint of DQ so M will be the midpoint of HD

To follow the proof, please draw a figure:

ReplyDeleteLet H’ be the orthocenter of ∆ADC

Let K, J be the feet of the ⊥s BH, DH’ on AC

Let KB, JE (both extended) meet at X

Let DJ(extended)meet circumcircle at Y

Join BY, DC. Also join XC, HY

We prove XHJD is a parallelogram

which would imply that the diagonals

XJ, HD bisect each other.

BH = DH’ (∵ each = 2 OZ, Z midpoint of AC)

C, D, F, J are cyclic.

∴ ∠FJD = ∠FCD = ∠BYD and so XJ ∥ BY

=> XBYJ is a parallelogram (∵XB, JY each ⊥ AC)

=> XB = JY = H’J (Property of orthocenter)

∴ XH = XB + BH = H’J + DH’ = DJ

Hence XHJD is a parallelogram, as required.

Refer Problem 729

ReplyDeletehttp://imageshack.us/photo/my-images/338/729gg.pdf/