NOTE THAT:
(Inappropriate answer writers shall be reported immediately)
In the figure given, QX and RX are bisectors of angles PQR and PRQ respectively of ∆PQR. If XS is perpendicular to QR and XT is perpendicular to PQ, prove that:
(i) ∆XTQ is congruent to ∆XSQ
(ii) PX bisects the angle P.
Attachments:
Answers
Answered by
16
(i) in figure,
angle XSQ + angle XSR = 180 ( linear pair)
so angle XSQ = 180 - 90
angle XSQ = 90
in triangle XTQ and triangle XSQ,
QX = QX ( common )
angle XTQ = angle XSQ ( each 90 degree)
angle TQX = angle XQS ( QX is bisector of angle PQR )
so triangle XTQ is congruent to triangle XSQ by angle side angle property.
hence proved
(ii) as angle TPX is equal to angle XPR
so we can say that PX bisects angle P
hope this helps you
please mark it as brainliest
angle XSQ + angle XSR = 180 ( linear pair)
so angle XSQ = 180 - 90
angle XSQ = 90
in triangle XTQ and triangle XSQ,
QX = QX ( common )
angle XTQ = angle XSQ ( each 90 degree)
angle TQX = angle XQS ( QX is bisector of angle PQR )
so triangle XTQ is congruent to triangle XSQ by angle side angle property.
hence proved
(ii) as angle TPX is equal to angle XPR
so we can say that PX bisects angle P
hope this helps you
please mark it as brainliest
stuti12:
so what do you think
by proving two congruent
and then PX = RX
and then by applying Pythagoras theorm we can prove the 2 part
I think that we have to first prove that we have to construct another perpendicular line from X to PR (let line on PR be Z)such that the 2 sides are equal of the base of that perp. line and TP and then prove the congruency of ∆XTP and ∆XZP.
* ignore the 1st line
Am I right?
yes
but don't report my answer
Stuti, since you are currently the first and the last writer, hence I would mark your answer as Brainiest.
Similar questions