Friday, February 7, 2014

Geometry Problem 975: Isosceles Right Triangle, 45 Degrees, Angle Bisector, Hypotenuse

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

Click the figure below to see the complete problem 975.

1. Let E be a point on AC such that DE⊥AC.

Then ΔABD congruent to ΔAED (AAS).
So AE=AB.

Now ΔEDC is an isosceles right triangle.
Thus EC=ED=BD.

Hence, AC=AE+EC=AB+BD.

2. Point E is on AB extended such that EB=BD. Then <BED=<DCA=45, so triangles ADE and ADC are congruent. Therefore AC=AE=AB+BE=AB+BD

3. AC=V2 . AB ; CD=V2. BD (angle bisect law)
BC=BD+CD= BD+V2.BD => BC+BD=2BD+V2.BD
BC=BD+V2.BD =>AC=V2.BC=V2.BD+2BD => AC=BC+BD

4. Extend AB to E such that AE = AC

Then Tr.s ABD and BCE are congruent ASA and so BD = BE

Hence AC = AE = AB + BE = AB + BD

Sumith Peiris
Moratuwa
Sri Lanka

1. Pure Geometry Solution 2

Extend DB to F such that FB = BD

Then Triangles AFB & ADB are congruent SAS and < FAB = 22.5

Hence < CFA = 180 - 45 - 67.5 = 67.5 and so AC = CF

Therefore AC = CF = BC + BF = AB + BD

Sumith Peiris
Moratuwa
Sri Lanka

Let AB=BC=a
AC=a*sqrt2
BD=atan22.5
AB+BD=a(1+tan22.5)=a[1+sqrt2-1]=a*sqrt2=AC