Squares ABDE and BCFG are drawn outside of triangle ABC : Prove that triangle ABC is isosceles if DG is parallel to AC.
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.
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.
No comments:
Post a Comment