Question
In the given figure, $AB || CD$. Prove that $P + q - r = 180.$
✓
Answer
Through $F$, draw $KH || AB || CD$
Now, $KF || CD$ and $FG$ is a transversal.
$\Rightarrow\angle\text{KFG}=\angle\text{FGD}=\text{r}^\circ (\text{i})$ [alternate angles]
Again $AE || KF$, and $EF$ is a transversal.
So, $\angle\text{AEF}+\angle\text{KFE}=180^\circ$
$\Rightarrow\angle\text{KFE}=180^\circ-\text{p}^\circ(\text{ii)}$
Adding $(i)$ and $(ii)$ we get,
$\angle\text{KFG}+\angle\text{KFE}=180-\text{p}+\text{r}$
$\Rightarrow\angle\text{EFG}=180-\text{p}+\text{r}$
$\Rightarrow\text{q}=180-\text{p}+\text{r}$ $\text{i}.\text{e}.,\ \text{p}+\text{q}-\text{r}=180$
Now, $KF || CD$ and $FG$ is a transversal.
$\Rightarrow\angle\text{KFG}=\angle\text{FGD}=\text{r}^\circ (\text{i})$ [alternate angles]
Again $AE || KF$, and $EF$ is a transversal.
So, $\angle\text{AEF}+\angle\text{KFE}=180^\circ$
$\Rightarrow\angle\text{KFE}=180^\circ-\text{p}^\circ(\text{ii)}$
Adding $(i)$ and $(ii)$ we get,
$\angle\text{KFG}+\angle\text{KFE}=180-\text{p}+\text{r}$
$\Rightarrow\angle\text{EFG}=180-\text{p}+\text{r}$
$\Rightarrow\text{q}=180-\text{p}+\text{r}$ $\text{i}.\text{e}.,\ \text{p}+\text{q}-\text{r}=180$
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