in the figure triangle ABC is right angled at B BD is perpendicular to AC prove that triangle ABC is similar to triangle ABC
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use concurrent condition
BD is common side
angle BDA =angle BDC
angle BAD=angle BCD
by angle side angle axiom
therefore triangle ABC is similar to triangle ABC
BD is common side
angle BDA =angle BDC
angle BAD=angle BCD
by angle side angle axiom
therefore triangle ABC is similar to triangle ABC
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