Question

In ∆PQR, O is the midpoint of side QR, seg OM is the bisector of angle POQ. Seg MN || side QR. Prove that angle MON = 90°​

Answer 1

Given :- In ∆PQR, O is the midpoint of side QR, seg OM is the bisector of angle POQ. Seg MN || side QR.

To Prove :- ∠MON = 90° .

Answer :-

MN ll QR,

→ PM / MQ = PN / NR { By BPT. }

also, OM is the bisector of angle POQ.

→ PO / OQ = PM / MQ { By angle bisector theorem.}

given that, OQ = OR , since O is mid - point of QR.

so,

→ PO / OR = PM / MQ

or,

→ PO / OR = PN / NR

then,

→ ON is also angle bisector of ∠POR .

now, let

• ∠POQ = 2a .
• ∠POR = 2b .

since, MN ll QR,

→ ∠PSN = ∠POR = 2b { corresponding angles.}

→ ∠PSM = ∠POQ = 2a { corresponding angles.}

then,

→ ∠PSN = ∠MSO = 2b { Vertically opposite angles.}

→ ∠PSM = ∠NSO = 2a { Vertically opposite angles.}

now, in ∆OMS ,

→ ∠OMS + ∠MSO + ∠MOS = 180° { By angle sum property.}

→ ∠OMS + 2b + a = 180°

→ ∠OMS = (180° - 2b - a) --------- Eqn.(1)

similarly, in ∆ONS ,

→ ∠ONS + ∠NSO + ∠NOS = 180° { By angle sum property.}

→ ∠ONS + 2a + b = 180°

→ ∠ONS = (180° - 2a - b) --------- Eqn.(2)

therefore, in ∆MON ,

→ ∠MON + ∠OMN + ∠MNO = 180° { By angle sum property.}

→ (a + b) + (180° - 2b - a) + (180° - 2a - b) = 180°

→ a + b - 2b - a - 2a - b + 360° = 180°

→ -2a - 2b + 360° = 180°

→ 2a + 2b = 360° - 180°

→ 2(a + b) = 180°

→ (a + b) = 90° .

hence,

→ ∠MON = (a + b) = 90° (Proved.)

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