The smaller of two consecutive even integers is five more than half of the greater.

< 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)

m∠abd = m∠dbc

step-by-step explanation:

given : ab bisects ∠dbc

as we know that bisection means ab divides   ∠dbc   in two equal angles named as   ∠dba and   ∠abc

so, m∠abd = m∠dbc  

because ab bisects ∠dbc.

hence , the first statement m∠abd = m∠dbc is true.

answer: i dont get it ur bad

step-by-step explanation: boi