Friday, October 28, 2022

Geometry Problem 1501: Square, Exterior Point, Congruent Angles, Segment, Measurement

Geometry Problem 1501. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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1. Let PG and AB intersect at point K.
Then Angle PKA= Angle PGD =Alpha.
Hence Triangle PKE is isosceles with PK=PE.
Extend FP to meet line DA at point L. We have isosceles Triangle PLH with PL=PH.
FL=PF+PL=PF+PH
Also FL=GK (since both segments make equal angle with side of square FL.sin(180- Alpha)=GK.sin(180-Alpha)=side of square)
We get PG=GK+PK=FL+PE=PF+PH+PE
16=4+5+PH, we get PH=7.

2. https://photos.app.goo.gl/JC82i2iPQsHZMMC26

Calculate PM= PG.sin (a)= 16.sin(a)
PN= PEsin(a)=5.sin(a)
PU=PF.sin(a)= 4 sin(a)
PV= PH.sin(a)
Since ABCD is a square => PU+PV=PM-PN
So PV= 7.sin(a) => PH= 7

3. Let FP, DH extended meet at V
Let PG, AB meet at U
Drop a perpendicular UX to CD, X on CD
Complete rectangle CDVW

PE = PU = 5

Tr.s VWF & GUX are congruent ASA
So FV = UG = 11
Hence PV = 11- 4 = 7

But PV = PH
Hence PH = 7

Sumith Peiris
Moratuwa
Sri Lanka