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Question
Which congruence criterion do you use in the following?
Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
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Solution
SSS, as the sides of ΔABC are equal to the sides of ΔDEF.
RELATED QUESTIONS
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of ∠A.
In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB = CE.
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.
If XS ⊥ QR and XT ⊥ PQ;
Prove that:
- ΔXTQ ≅ ΔXSQ.
- PX bisects angle P.
In the following diagram, ABCD is a square and APB is an equilateral triangle.
- Prove that: ΔAPD ≅ ΔBPC
- Find the angles of ΔDPC.
In the following figure, OA = OC and AB = BC.
Prove that: ΔAOD≅ ΔCOD
In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced up to point R so that CR = BP.
Prove that QR bisects PC.
Hint: ( Show that ∆ QBP is equilateral
⇒ BP = PQ, but BP = CR
⇒ PQ = CR ⇒ ∆ QPM ≅ ∆ RCM ).
