## Monday, May 19, 2008

### Elearn Geometry Problem 60

See complete Problem 60
Isosceles triangle and angles. Level: High School, SAT Prep, College geometry

1. use sine rule for both the triangles and u may get answer

2. Let P be a point such that triangle BPC is equilateral, then triangles CAB and DBP are congruent by LAL, because this, triangle DPC is an 80-80-20.

Therefore, x=20°

Quod Erat Demonstrandum.

3. Or, we can let E be the point on the opposite side of AC as B such that triangle ACE is congruent to triangle BDC, where angle CAE=10. Then BAE is isoceles, so angle CBE=10. Also, CD=CE, so B must lie on the circumcircle of triangle BDE as angle DBC and angle CBE are equal and intercept equal arcs of this circle. Thus angle DCE=160, so
160+(70-x)+(170-x)=360, so x=20.

4. Let E be such that EB = AB and < EBA = 20

Then Tr. EBC is equilateral and Tr. ABE is isoceles Tr.s ACE & Tr, BDC Tr. are congruent and x = 20

Sumith Peiris
Moratuwa
Sri Lanka

5. Let E inside the triangle such that ACE be an equilateral triangle; see that triangles BCE and CBD are congruent, s.a.s., so <BCD=<CBE=20 degs.

A Happy New Year 2016!!