Advertisements
Advertisements
Question
D, E, F are the mid-point of the sides BC, CA and AB respectively of ΔABC. Then ΔDEF is congruent to triangle
Options
ABC
AEF
BFD, CDE
AFE, BFD, CDE
Advertisements
Solution
It is given that D, E and Fare the mid points of the sides BC , CA and AB respectively of ΔABC
FE =BD (By mid point theorem)
BD = DC (As it is mid point)
Now in ΔAFE and ΔDFE
FE(Common)
DF = AE (Mid point)
AF = DE (Mid point)
⇒ ΔFED ≅ ΔBFD
⇒ ΔDFE ≅ ΔDCE
Hence (d)
ΔDFE ≅ AFE
≅ BFD
≅ CDE
APPEARS IN
RELATED QUESTIONS
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?
In the figure, the two triangles are congruent.
The corresponding parts are marked. We can write ΔRAT ≅ ?
In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C = 30° and ∠D = 90°. Are two triangles congruent?
If the following pair of the triangle is congruent? state the condition of congruency:
In ΔABC and ΔPQR, AB = PQ, AC = PR, and BC = QR.
The following figure shows a circle with center O.
If OP is perpendicular to AB, prove that AP = BP.
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD.
Prove that :
(i) ΔABD and ΔECD are congruent.
(ii) AB = CE.
(iii) AB is parallel to EC
In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that : ED = EF
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O.
Prove that : (i) BO = CO
(ii) AO bisects angle BAC.
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
