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

Labels:
area,
congruence,
right triangle,
similarity,
trapezoid

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http://img442.imageshack.us/img442/5306/problem8841.png

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