Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 884.
Friday, June 7, 2013
Problem 884: The Pythagorean Curiosity, Fifteen Conclusions
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ReplyDeletePartial answer to the problem.
Questions 2, 5, 6 .
note that ∆RCQ and ∆MXQ are congruent ( case SAS)
So MQ=QR and ∠ (CQR)= ∠ (MQX)=y
In isosceles triangle MQH we have ∠MQH)+2x=180
At point Q we have ∠ (MQH)+2y+180=360 => ∠ (MQH)+2y=180
So x=y and MH//XQ//AC
Similarly DE//PY//BA
Let Q’ is the projection of Q over QC and P’ is the projection of P over NB
We have QQ’=CQ.sin(C )=b.c/a and PP’=PB.sin(B)=c.b/a
So QQ’=PP’
Note that QQ’=GG’ and PP’=FF’ ( congruent triangles)
So FG//NR//BC
Questions 9, 12, 13
Note that right triangles XAY, MX’X and YY’D are congruent ( case HA)
So MX’=AX= b => MH=4b
So A’X’AY’ is a square and AA’ is a diagonal so AA’ is a bisector of angle A and A’ .
AA’ ⊥ PQ
Hi peter,
Deletei have problems with proof MH//XQ//AC when the first triangle construct is not right triangle , but a scalelane triangle , can u help it
If triangle ABC is a scalene triangle , you can follow the similar logic as my solution above to prove that MH//XQ//AC . You can email me the sketch and details of the problem. I will trỵ
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