Question
Line segments AB and CD intersect at O such that AC perpendicular DB. It $\angle \text{CAB}=35^\circ$ and$ \angle \text{CDB}=55^\circ$. Find $\angle \text{BOD}$
✓
Answer
We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.
$\angle \text{CAB}=\angle \text{DBA}$ (Alternate interior angles)
$\angle \text{DBA}=35^\circ$
We also know that the sum of all three angles of a triangle is 180°
Hence, for $\triangle \text{OBD},$ we can say that:
$\angle \text{DBO}+\angle \text{ODB}+\angle \text{BOD}=180^\circ$
$35^\circ+55^\circ+\angle \text{BOD}=180^\circ$ $(\angle \text{DBO}=\angle \text{DBA}$ and $\angle \text{ODB}=\angle \text{CDM})$
$\angle \text{BOD}=180^\circ-90^\circ$
$\angle \text{BOD}=90^\circ$
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