Question 11
Question: In D ABC, AD is a median of BC. Prove that AB+AC>2AD.
Answer: Given:
AD is a median of D ABC. D is the midpoint of BC
To prove:
AB+AC>2AD
Construction:
Produce AD to E such that AD=DE. Join CE.
Proof:
In triangles ABD and CED,
AD = DE ( by construction)
BD=DC ( D is the midpoint of BC)
(given)
( vertically opposite angles)
In D ACE,
AC+CE>AE (sum of any two sides of a
triangle is greater than the
third side)
AC+AB>AE (Substituting AB for CE proved above)
But AE=AD+DE ( from the figure)
=AD+AD
=2AD
DE=AD (by construction)
Question 12
Question: In the adjoining figure PQ=PS. Prove that PS
Answer: Given:
PQ=PS
To prove:
PSProof:
,
Question 13
Question: In the adjoining figure AB>AC. Show that AB>AD.
Answer: Given:
AB>AC
To prove:
AB>AD
Proof:
AB>AC (given)
(Converse of theorem on inequality)
In D ABD,
Question 14
Question: Show that in a right triangle the hypotenuse is the longest side.
Answer: Given:
In right triangle AC is the hypotenuse.
To prove:
AC is the longest side.
Proof:
In a triangle there can be only one right angle.
AC>BC ...(i)
AC>AB ...(ii)
From (i) and (ii), AC is the longest side.
Question 15
Question: In the adjoining figure PQRS is a quadrilateral. PQ is the longest side and RS is its shortest side. Prove that 
.
Answer: Given:
PQRS is a quadrilateral.
PQ is the longest side.
RS is the shortest side.
To prove:
Construction:
Proof:
In D RPQ,
PQ>QR
PQ is the greatest side. (given)
In D SRP,
PS>SR
RS is the shortest side.
(i) + (ii)
In D PQS,PQ > PS (given)
PQ is the longest side.
In D SRQ, SR
SR is the shortest side.
By adding (iii) and (iv), we get
Question 16
Question: 
Answer: Given:
To prove:
AB>BD
Proof:
(Exterior angle is > interior opposite angle)
AB>BD (theorem on inequality)
Question 17
Question: PQRS is a quadrilateral. Prove that
(i) PQ+QR+RS+SP>PR+QS
(ii) PQ+QR+RS+SP<2(PR+QS)
Answer: Given:
PQRS is a quadrilateral. Diagonals PR and QS intersect at O.
To prove:
(i) PQ+QR+RS+SP>PR+QS
(ii) PQ+QR+RS+SP<2(PR+QS)
Proof:
In D PSR,
PS+SR>PR ...(i)
(Sum of any two sides > third side)
In DPQS,
PS+PQ>QS ...(ii)
(Sum of any two sides > third side)
In DQRS,
QR+SR>QS ...(iii)
(Sum of any two sides > third side)
InDPQR,
PQ+QR>PR ...(iv)
Add (i), (ii), (iii) and (iv), we get
2(PQ+QR+SR+OS)>2(PR+QS) ...(v)
PQ+QR+SR+PS>PR+QS (divide both sides by 2)
PO + OQ > PQ
OQ+OR >QR
OR + OS > SR
OS + OP > SP
By adding,
PQ+QR+SR+PS<2(PR+QS).
Question 18
Question: In the adjoining figure AB=AC. Prove that BD>CD.
Answer: Given:
AB=AC
To prove:
BD>CD
Proof:
AB=AC ( given )
(angles opposite to equal sides are equal)
(side opposite to greater angle)
