Question

AB is a line segment. AX and BY are two equal line segments drawn on the opposite sides of AB such that AX||BY. If AB and XY intersect each other at P, prove that (i) triangle APX is congruent to triangle BPY and (ii) AB and XY bisected each other.

Answer 1

i) In triangle APX and BPY
Angle APX = Angle BPY (Vertically opposite angles)
AX||BY and XY is transversal
It implies Angle AXP=Angle BYP
AX=BY
Triangle APX Congruent to Triangle BPY

PX = PY (BY CPCT)
PA = PB (BY CPCT)
It implies AB and XY bisect each other

Answer 2

Step-by-step explanation:

Since AX||BY and AB is transversal.

so, angle BAX = angle ABY. { Alter. Int. Angles}

similarly, angle AXY = angle BYX. { Transversal XY intersects parallel lines AX and BY at X and Y}

If AX and YB and AB is transversal

so, angle A and angle B is alter. int. angles

In ∆ PAX and ∆ PBY

angle A = angle B. {alt. int. ang.}

AX = BY. { given}

angle XPA = angle BPY. { vert. opp. ang.}

∆PAX is congurent to ∆PBY {angle side angle}

AP = BP. {c.p.c.t}

PX = PY {c.p.c.t}

Hence, ∆APX is congurent to ∆ BPY and AB and XY bisect each other.