Question

In the figure, PQ=PR and QA=BR. Prove that
I) triangle PQA is congruent to triangle PRB
II) angle PAB = angle PBA

Answer 1

Answer:

Proved:

Step-by-step explanation:

Given parameters:

• PQ=PR
• QA=BR

Prove that

I) triangle PQA is congruent to triangle PRB

As PQ = PR,<PQR = < PRQ (equal sides' opposite angles are equal)

In ∆ PAQ and ∆PBR,

• PQ = PR (Given)
• <PQR = < PRQ (See above)
• QA = BR (Given)

By SAS rule of congruency, ∆PAQ is congruent to ∆PBR

By c.p.c.t.c. <PAQ = <PBR ...(i)

From equation (i),Subtract 180° from both sides:

• 180° - <PAQ = 180° - <PBR ...(ii)

As PR is a straight line,and AP and BP are lines dividing the angles on the line, we can apply linear pair theorem that is :

• Angles formed by the line segment from any outer point onto a line equals 180°, provided the line is straight.
• <PAQ + <PAB = 180° for segment AP, thus <PAB = 180°-<PAQ ...iii
• <PBA+<PBR = 180° for segment BP thus <PBA = 180° - <PBR ...iv

From (ii) we can rewrite equation (iii) and (iv) as

• <PBA = <PBR
• <PAQ = <PAB

Hence Proved angle PAB = angle PBA