在四边形 ABCD 中，BC∥AD,CA 平分∠BCD,O 为对角线的交点，CD = AO ，BC = OD ，
求 ∠ABC 的度数

Question

在四边形 ABCD 中，BC∥AD,CA 平分∠BCD,O 为对角线的交点，CD = AO ，BC = OD ，
求 ∠ABC 的度数

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given that in quadrilateral ABCD, $BC \parallel AD$, $CA$ bisects $\angle BCD$, $O$ is the intersection point of the diagonals, $CD = AO$, and $BC = OD$. We need to find the measure of $\angle ABC$. ‖ ‖ step 2 ⋮ Since $CA$ bisects $\angle BCD$, we have $\angle BCA = \angle DCA$. ‖ ‖ step 3 ⋮ Given $CD = AO$ and $BC = OD$, triangles $BCD$ and $AOD$ are congruent by the Side-Angle-Side (SAS) criterion. ‖ ‖ step 4 ⋮ Since triangles $BCD$ and $AOD$ are congruent, $\angle BCD = \angle AOD$. ‖ ‖ step 5 ⋮ Since $BC \parallel AD$, $\angle ABC = \angle AOD$ (corresponding angles). ‖ ‖ step 6 ⋮ Therefore, $\angle ABC = \angle BCD / 2$. ‖ ‖ step 7 ⋮ Since $\angle BCD$ is bisected by $CA$, $\angle BCD = 2 imes \angle BCA$. ‖ ‖ step 8 ⋮ Let $\angle BCA = x$. Then $\angle BCD = 2x$ and $\angle ABC = x$. ‖ ‖ step 9 ⋮ Since $\angle BCD$ is an exterior angle of $ riangle ABC$, $\angle BCD = \angle ABC + \angle BAC$. ‖ ‖ step 10 ⋮ Therefore, $2x = x + \angle BAC$. ‖ ‖ step 11 ⋮ Solving for $\angle BAC$, we get $\angle BAC = x$. ‖ ‖ step 12 ⋮ Since $\angle BAC = x$ and $\angle ABC = x$, $\angle ABC = 45^\circ$. ‖ ∻Answer∻ ⚹ $\angle ABC = 45^\circ$ ⚹ ∻Key Concept∻ ⚹Angle Bisector and Parallel Lines⚹ ∻Explanation∻ ⚹The problem uses properties of angle bisectors and parallel lines to determine the measure of $\angle ABC$.⚹◊What is the relationship between the angles in a quadrilateral where BC is parallel to AD, CA bisects angle BCD, and CD = AO?⍭ Generate me a similar question◊

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