Plzz solve1,2.... URGENT

Plzz solve1,2.... URGENT Trian 1. In the given figure, it is given that BC- CE and ABG DEF. is congruent to OCE 2. In the given figure. P Is a point in the interior of L AOB. is perpendicularto OA and PM is perpendicular to 0B such that PL 2M, show that op is the bisector o' z AOB.

Dear Student,

Please find below the answers to the questions.

Answer 1:

It is given in the question that $\angle ABG = \angle DEF.....\left(i\right)$
As,

$$\begin{gathered} \angle GBC + \angle ABG = 180 and \angle DEC + \angle DEF = 180 (angles on a straight line) \\ or, \\ \angle ABG = 180 - \angle GBC and \\ \angle DEF = 180 - \angle DEC \end{gathered}$$

Thus replacing the values in equation (i), we get,

$$\begin{gathered} 180 - \angle GBC = 180 - \angle DEC \\ or, \\ \angle GBC = \angle DEC.....\left(ii\right) \end{gathered}$$

Thus, considering $△GCB$ and $△DCE$, we can see that,
$$\begin{gathered} \angle GCB = \angle DCE \left\{Vertically opposite angles\right\} \\ BC = CE \left\{Given in the question\right\} \\ \angle GBC = \angle DEC \left\{From \left(ii\right)\right\} \end{gathered}$$

Hence, based on ASA (Angle Side Angle) property we can say that $△GCB$ and $△DCE$ are congruent with each other.

Answer 2:

In right $△OLP$, we can see that,
$O{L}^2+P{L}^2=O{P}^2.....\left(i\right) (U\sin g Pythagorous theorem i.e. Perpendicula{r}^2+Bas{e}^2=Hypotenus{e}^2)$
And, In right $△OMP$, we can see that,
$O{M}^2+P{M}^2=O{P}^2.....\left(ii\right) (U\sin g Pythagorous theorem i.e. Perpendicula{r}^2+Bas{e}^2=Hypotenus{e}^2)$
Thus, from (i) and (ii), we can see that,

$$\begin{gathered} O{L}^2+P{L}^2=O{M}^2+P{M}^2 \\ or, \\ O{L}^2=O{M}^2 (Since PL=PM given in the question) \\ or, \\ OL=OM......\left(iii\right) \end{gathered}$$

Considering $△OLP$ and $△OMP$, we can see that;
$$\begin{gathered} PL = PM \left\{Given in the question\right\} \\ \angle PLO = \angle PMO \left\{Perpendiculars, given in the question\right\} \\ OL = OM \left\{From \left(iii\right)\right\} \end{gathered}$$

Hence, based on SAS (Side Angle Side) property we can say that $△OLP$ and $△OMP$ are congruent with each other.

Therefore, $\angle LOP = \angle MOP (Angles opposide to the equal and congruent sides are always equal and congruent)$
Since,
 
$$\begin{gathered} \angle AOB = \angle LOP + \angle MOP \\ and, \\ \angle LOP = \angle MOP \end{gathered}$$

Thus, we can definitely say that $OP$ is the bisector of $\angle AOB$.

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