Question

AB is trisected at P and Q such that P is between A and Q and AL perpendicular to AB, BM perpendicular to AB. If LQ and MP intersect at O and LQ= MP prove that ∆OPQ is isosceles​

Answer 1

Answer:

Given P is equidistant from points A and B

PA=PB

and Q is equidistant from points A and B

QA=QB

In PAQ and PBQ

AP=BP from

AQ=BQ from

PQ=PQ (common)

So, PAQPBQ (SSS congruence)

Hence ∠APQ=∠BPQ by CPCT

In PAC and PBC

AP=BP from

∠APC=∠BPC from

PC=PC (common)

PACPBC (SAS congruence)

AC=BC by CPCT

and ∠ACP=∠BCP by CPCT

Since, AB is a line segment,

∠ACP+∠BCP=180

(linear pair)

∠ACP+∠ACP=180

from (4)

2∠ACP=180

∠ACP=

2

180

=90

Answer 2

In ∆ LAQ and ∆ MBP

®LQ = MP (given)

®Angle LAQ = Angle MBP (90°)

®AQ = BP (AB = 3AP or 3PQ or 3QB)

By RHS congruence, ∆ LAQ =~∆ MBP

i.e. Angle AQL = Angle BPM. ----(1)

Consider in ∆ OPQ,

Angle OQP = Angle QPO

(Angle OQP = Angle AOL & Angle QPO = Angle BPM)

i.e. OP = OQ

Thus OPQ is an isosceles triangle.