Friday, June 1, 2012

Squares on Sides of a Triangle

Squares ABDE and BCFG are drawn outside of triangle ABC : Prove that triangle ABC is isosceles if DG is parallel to AC.
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Given that DG is parallel to AC. Draw a perpendicular from B to AC , this is also perpendicular to DG. Let the perpendicular intersect AC at P and DG at Q.

Since ∠ABP = 90° - ∠DBQ = ∠BDQ and AB = BD , the right triangles ABP and BDQ are congruent (by ASA Criteria) , hence AP = BQ (by CPCT).

Similarly , right triangles CBP and BGQ are congruent and BQ = PC. So , by the above , AP = CP and BP is perpendicular to AC , this implies that AB = BC , hence triangle ABC is isosceles.

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