Proposed Problem

See also: Complete Problem 281, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Friday, April 17, 2009

### Problem 281: Triangles, Angles, Isosceles, Equilateral, Congruence

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Wonder if this problem has purely geometry proof. I solved it using trigonometry.

ReplyDeleteExtend AD, from C draw a segment,intersect with Extended AD at H, CH= AC, tri ACH is isosceles;

ReplyDeleteconnect BH;

<BDC = 150-alfa, so <HDC = 90-alfa, so <BDH =60;

because <ACD + <DCH + 4 alfa =180 =90-3alfa +<DCH + 4alfa ; so <DCH =90-alfa;so DHC is isosceles;

So AC=CH=DH; so DH =DB, so BDH is isosceles; and <BDH is 60, so Tri BDH is equilateral tri,;so BH=DH=HC ; so use H as center, radius =HC, make a circle connect BDC.

because <BDH=60, so X=30

Extend AD to E such that AC = CE

ReplyDelete< EDC = 90 - @ and < AEC = 2@ so DE = EC

Now < BDE = 60 so BDE is equilateral and E is the centre of BCD.

Hence x = 1/2 < BED = 1/2 X 60 = 30

We need to observe that x = 30 irrespective of what value @ holds

Sumith Peiris

Moratuwa

Sri Lanka