Question

LMNO is a rhombus, seg OP is perpendicular to side LM and
seg OQ
is perpendicular to side MN. Prove that
seg OP is congruent to seg OQ
PLZZ ans​

Answer 1

Given,

∠OQN=∠OPL

LMNO is a rhombus

so, LM||ON and OL||MN

Then,

∠ONQ=∠LMN -( i ) ( NO||LM QM is Transversal )

( angles on side of Transversa)

also,

∠LMN=∠PLO -( ii ) (OL||MN PM is Transversal )

( angles on side of Transversa)

then

∠ONQ= ∠PLO --( iii )

in ∆OPL

∠POL+∠OPL+∠PLO=180 ( angles sum property )

∠POL=180-∠OPL-∠PLO

∠POL=180-∠OQN-∠ONQ --( iv ) (from above)

in ∆OQN

∠NOQ+∠OQN+∠ONQ=180( angles sum property )

∠NOQ=180-∠OQN-∠ONQ

∠NOQ=∠POL (from iv )

consider ∆OPL and ∆OQN

∠OPL=∠OQN ( given )

∠POL=∠NOQ (from iii )

∠PLO=∠ONQ (from iv )

therefore ∆OPL≅∆OQN

from congruence OP=OQ