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

Statement Reason

∆ABC and ∆ADC are isosceles Given

Segment AC is a common base

to ∆ABC and ∆ADC Given

Let K be the midpoint of the base

of segment AC, then BK is the

perpendicular bisector of segment perpendicular bisector of isosceles AC. triangle property.

Also, DK is the perpendicular

bisector of segment AC. perpendicular bisector of isosceles

triangle property

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< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)