Proposed Problem

Click the figure below to see the complete problem 408 about Cyclic quadrilateral, Perpendicular, Parallelogram, Congruence.

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Complete Problem 408

Level: High School, SAT Prep, College geometry

## Friday, December 25, 2009

### Problem 408: Cyclic quadrilateral, Perpendicular, Parallelogram, Congruence

Labels:
circle,
congruence,
cyclic quadrilateral,
parallelogram,
perpendicular

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extend BF to P, and CG to Q ( P and Q on circle )

ReplyDeletea) ang NBC = ang FPQ ( BP//CQ, and BPQC cyclic) (1)

b) join A to Q, ang ADE = ang AQG ( have same arc ABC)

=> ang EAG = ang GAQ

=> GQ = GM (2)

c) at the same way

=> NF = FP (3)

d) from (2) and (3)

ang MNF = ang FPQ (4)

e) from (1) and (4)

ang NBC = ang FNM

=> BC//NM

can you tell me, please

ReplyDeletewhy my solution is deleted

Accidentally deleted.

ReplyDeleteFollowing the solution posted by c.t.e.o on December 27, 2009.

extend BF to P, and CG to Q ( P and Q on circle )

a) ang NBC = ang FPQ ( BP//CQ, and BPQC cyclic) (1)

b) join A to Q, ang ADE = ang AQG ( have same arc ABC)

=> ang EAG = ang GAQ

=> GQ = GM (2)

c) at the same way

=> NF = FP (3)

d) from (2) and (3)

ang MNF = ang FPQ (4)

e) from (1) and (4)

ang NBC = ang FNM

=> BC//NM

thanks

ReplyDeletethe main idea of solution was

finding AD as simetrical line

=>

everything under it, is congruent to everything above it

( segment, angle, triangle, ... )

Problem 408

ReplyDeleteThe points M, N are orthocenter of triangles ACD, ABD respectively. Bring OP perpendicular AD.Then BN=//2OP and CM=//2OP. So BN=//CM, therefore BCMN is parallelogram.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE