Question

prov that the line segment joining the midpoint of any two sides of a triangle is parallel to the third side and equal to half of it

Answer 1

I hope this will be helpful

Answer 2

The figure will be something like this....♂

Given: It is given that AQ is equal to CQ and AP is equal to PB.

Construction: (i) ∆ABC is given and we have to draw a line CD parallel to line AD.

(ii) Extend the line PQ to line CD such that PQ will meet CD at R.

To prove: (i) PQ is parallel to BC
(ii) PQ =1/2BC

Proof: ∆APQ and ∆CRQ:
angle AQP= angle CQR (Vert. opp. angles)
AQ = CQ (given)
angle PAQ = angle RCQ (Alt. int. angles)
So, ∆APQ is congruent to ∆CRQ (ASA)
Hence, AP=CR and PQ=RQ (both by CPCT)

AP = PB and, (1)
AP = CR,(2)
So, from (1) &(2): PB = CR

Hence, PBCR is a parallelogram (because opp. sidws are equal and parallel)
So, PQ is parallel to BC. {Hence (i) proved}

PR =BC (opp.sides of paral. are equal)

PQ = QR (we proved it above by CPCT)---(3)
and, PQ+QR=PR----(4)
PR =BC----(5)
From (3),(4)&(5):
2PQ=BC
So, PQ=1/2 BC {Hence (ii) proved}


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