Advertisements
Advertisements
प्रश्न
In a parallelogram ABCD, E is the midpoint of AB and DE bisects angle D. Prove that: BC = BE.
Advertisements
उत्तर
∠CDE = ∠DEA ...(ALternate angles)
∠CDE = ∠EDA ...(Given DE bisects ∠D)
∠DEA = ∠EDA
⇒ AD = AE ....(i)(Sides opposite to equal angles are equal)
Now, AD = BE ....(ii)(Opposite to equal angles are equal)
And, AE = BE ....(iii)(E is the mid-point of AB)
⇒ BC = BE. ....(proved)
APPEARS IN
संबंधित प्रश्न
The alongside figure shows a parallelogram ABCD in which AE = EF = FC.
Prove that:
- DE is parallel to FB
- DE = FB
- DEBF is a parallelogram.
The following figure shows a trapezium ABCD in which AB is parallel to DC and AD = BC. 
Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.
Points M and N are taken on the diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram.
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: RT bisects angle R
ABCD is a parallelogram. The bisector of ∠BAD meets DC at P, and AD is half of AB.
Prove that: BP bisects ∠ABC.
In a parallelogram ABCD, E is the midpoint of AB and DE bisects angle D. Prove that:CE is the bisector of angle C and angle DEC is a right angle
In the given figure, the perimeter of parallelogram PQRS is 42 cm. Find the lengths of PQ and PS.
Find the perimeter of the parallelogram PQRS.
In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO is `9/8` of the area of the parallelogram ABCD
In the following figure, ABCD and AEFG are two parallelograms. If ∠C = 55º, determine ∠F.
