From the following figure, prove that:
AB>CD
\text { In a triangle } P Q R ; Q R=P R \text { and } \angle P=360 \text { . Which is the largest side of the triangle? }
If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a
cm and b cm. Find the values of a and b such that a is less than b.
In each of the following figures write BC, AC, and CD in ascending order of their lengths.
(i)
Arrange the sides of triangle BOC in descending order of their lengths. BO and CO are
bisectors of angles ABC and ACB respectively.
D is a point in side BC of triangle ABC. If AD>AC, show that AB>AC.
In the following figure, BAC = 600 and ABC = 650.
Prove that:
(i) CF > AF
(ii) DC > DF
From the following figure; prove that:
(i) AB > BD
(ii) AC > CD
(iii) AB + AC > BC
In a quadrilateral ABCD; prove that:
(i) AB + BC + CD > DA
(ii) AB + BC + CD + DA > 2AC
(iii) AB + BC + CD + DA > 2BD
In the following figure, ABC is an equilateral triangle and P is any point in AC; prove that:
(i) BP > PA
(ii) BP > PC
\mathrm{P} \text { is any point inside the triangle } \mathrm{ABC} . \text { Prove that: } \angle \mathrm{BPC}>\angle \mathrm{BAC} \text { . }
\text { In triangle } \mathrm{ABC} ; \mathrm{AB}=\mathrm{AC} \text { and } \angle \mathrm{A}: \angle \mathrm{B}=85 ; \text { find } \angle \mathrm{A} \text { . }
Prove that the straight line joining the vertex of an isosceles triangle to any point in the
the base is smaller than either of the equal sides of the triangle.
In the following diagram; AD = AB and AE bisects angle A. Prove that:
(i) BE = DE
(\text { ii) } \quad \angle \mathrm{ABD}>\angle \mathrm{C}
The sides AB and AC of a triangle ABC are produced; and the bisectors of the external
angles at B and C meet at P. Prove that if AB>AC, then PC > PB.
In the following figure; AB is the largest side and BC is the smallest side of triangle ABC.
In quadrilateral ABCD, side AB is the longest, and side DC is the shortest. Prove that:
\text { (i) } \angle \mathrm{C}>\angle \mathrm{A}
\text { (ii) } \angle \mathbf{D}>\angle \mathbf{B}
In triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets
the opposite side at point D, prove that: \angle \mathrm{ADC} \text { is greater than } \angle \mathrm{ADB}
An isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point
E lies on BC produced (BC being produced through vertex C), prove that:
(i) AC > AD
(ii) AE > AC
(iii) AE > AD
Given: ED = EC
Prove: AB + AD > BC
In triangle ABC, AB > AC and D is a point in side BC. Show that: AB>AD.
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