PLEASE HELP FOR BRAINIEST ASAP!!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.

Accepted Solution

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