Question

give answer of this question asked in figure

Answer 1

$\textbf{ Hello Mate! }$

Assume that PQ || BC and Sum of interior angles on same side of transversal BQ intersecting parallel lines PQ and BC

Then, /_PQB + /_ QBC = 180°
/_PQB = 180° - 90°
/_ PQB = 90°

(i) Hence, PQ is perpendicular at AB

Then, /_ QPB = /_ PBC [ Alternate Angles ]

In ∆ CPB and ∆ APB, we have

BP = BP [ Common ]
AP = PC [ P is midpoint on AC ]
/_CBP = /_ APB [ BP bisects AC

Hence, both ∆s are congruent by SAS congruency.

PB = PA [ c.p.c.t ]

As PA is half of AC
(iii) PA = PB = 1/2 ( AC )

In ∆ PQB and ∆ PAQ we habe

PQ = PQ [ common ]
PB = PA [ proved above ]
/_PQB = /_ PQA [ 90° each ]

Hence both ∆s are congruent by RHS congruency.

So, AQ = BQ [ c.p.c.t ]

(ii) Hence, Q is mid point of AB

Have great future ahead!