'Tri' means

Two circles are congruent.If the radius of one circle is 3cm, what is the radius of the other circle?

Two circles are congruent.If the radius of one circle is 1cm, then the diameter of the other circle is

ΔABC ≅ ΔPQR.If AB = 5cm, ㄥB = 40⁰and ㄥA = 80⁰, then which of the following is true?

In two triangles ABC and DEF, ㄥA = ㄥD, ㄥB = ㄥE and AB = EF, then are the two triangles congruent? If yes, by which congruency rule?

In a ΔDEF, if ㄥD = 30°, ㄥE = 60° then which side of the triangle is longest and which side is shortest?

In ΔABC, ㄥA = 60°, ㄥB = 40°, which side of this triangle is the smallest? Give reasons for your answer.

In the figure below, ABCD is a square and P is the mid-point of AD. BP and CP are joined. Prove that \(\angle\)PCB = \(\angle\)PBC.

In figure \(\angle\)B = \(\angle\)E, BD = CE and \(\angle\)1 =\(\angle\)2. Show \(\triangle ABC\cong \triangle AED.\)

In the figure, given AC > AB and AD is the bisector of \(\angle\)A. Show that \(\angle\)ADC > \(\angle\)ADB.

In the figure, ABC is an isosceles triangle in which AB = AC and LM is parallel 10 BC If LA = 50°, find \(\angle\)LMC.

In the given figure, AB = AC and BE and CF are bisectors of \(\angle\) B and \(\angle\)C respecively. Prove that  \(\triangle\)EBC= \(\triangle\)FCB

In a \(\triangle\)ABC, AD \({ \bot }\) BC, BE\({ \bot }\)AC and CF\({ \bot }\)AB. Prove that AD + BE + CF < AB + BC + CA.

In the given figure, PQR is a triangle and S is any point in its interior. Show that SQ + SR < PQ + PR.

In the figure, prove that CD + DA + AB + BC > 2AC.

In a rectangle ABCD, E is a point which bisects BC Prove that AE = ED

In figure OA = OB, OC = OD and \(\angle AOB=\angle COD\). Prove that AC = BD

In the given figure, AB and CD are perpendicular to the line segment AD. AD and BC intersect at P such that PA = PD. Prove that:
(i) AB = CD
(ii) P is the mid-point of BC.

Prove that the medians of an equilateral triangle are equal.

In the given figure, AD is the bisector of \(\angle\)BAC and \(\angle\)CPD = \(\angle\)BPD. Prove that\(\triangle CAP\cong \triangle BAP\) and CP = BP.