Advertisements
Advertisements
प्रश्न
D, E, F are the mid-point of the sides BC, CA and AB respectively of ΔABC. Then ΔDEF is congruent to triangle
विकल्प
ABC
AEF
BFD, CDE
AFE, BFD, CDE
Advertisements
उत्तर
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
संबंधित प्रश्न
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 the figure, the two triangles are congruent.
The corresponding parts are marked. We can write ΔRAT ≅ ?

If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
If the following pair of the triangle is congruent? state the condition of congruency:
In ΔABC and ΔQRP, AB = QR, ∠B = ∠R and ∠C = P.
In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that: BD = CD

In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles.
Prove that: XA = YC.
In the following figure, OA = OC and AB = BC.
Prove that: ΔAOD≅ ΔCOD
In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°.
Prove that AD = FC.
AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.
