Question

pls solve this guys ​

Answer 1

Answer:

In triangle POL and triangle POM

Angle PLO = Angle PMO ( each 90)

OP = OP (common)

PL = PM (given)

So triangle POL in congruent to triangle POM by (RHS)

Step-by-step explanation:

triangles are congruent by RHS (RIGHT ANGLE- HYPOTENUSE - SIDE) criterion of congruency.

Answer 2

$\huge\mathbb{SOLUTION}$

Given :-

PL= PM

$\sf PM \perp OB \\ \sf PL\perp OA$

To prove :-

$\sf\triangle PLO\cong\triangle PLO$

Proof :-.

In ∆PLO AND ∆PMO

PL=PM (Given )

$\sf\angle PMO=\angle PLO =90^{\circ}$ (Given)

PO =PO (common side)

By RHS criteria $\sf\triangle PMO\cong \triangle PLO$

$\large\mathtt{\red{SOME\: INFORMATION\: Related\: triangle}}$

SAS :- which has Two sides and one angle

ASA :- Which has two angle and one side

SSS :-Which has three sides

RHS :- Right hypotenuse side