Question

Math

Posted about 2 months ago

It gives me a wrong answer, because the standard answer is 126

```
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
```

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Answer from Sia

Posted about 2 months ago

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 $CA$, $\angle BCD = 2 \times \angle BCA$

step 8

Let $\angle BCA = x$. Then $\angle BCD = 2x$ and $\angle ABC = x$

step 9

Since $\angle BCD$ is an exterior angle of $\triangle ABC$, $\angle BCD = \angle ABC + \angle BAC$

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 $\angle ABC$.

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