Draw two isosceles triangles, ∆ABC and ∆ADC with common base AC. Vertexes B and D are in the opposite semi-planes determined by AC. Draw the line segment BD. Prove: AC ⊥ BD.
suppose AC and BD intersect at E ΔABC is isosceles triangle Given AB=CB Definition of isosceles triangle ΔADC is isosceles triangle Given AD=CD Definition of isosceles triangle BD=BD Reflexive property ΔBAD=ΔBCD SSS theorem ∠ABE=∠CBE CPCTC ∠BAE=∠BCE Definition of isosceles triangle ΔABE=ΔCBE ASA theorem ∠AEB=∠CEB CPCTC
∠AEB+∠CEB =180 Linear pair of angles are supplementary ∠AEB=∠CEB =90 Algebra
∠AEB is a right triangle Definition of right triangle AC ⊥ BD Definition of perpendicular
