Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर
Since BC = BE
⇒ ∠BEC = ∠BCE ...(Angles opposite to equal sides are equal)
∠BEC = ∠ECD ...(Alternate angles)
⇒ ∠BCE = ∠ECD
⇒ CE is the bisector of ∠C ....(proved)
∠DCE = `(1)/(2)∠"C"` ...(Given CE bisects ∠D)
∠CDE = `(1)/(2)∠"D"` ...(Given DE bisects ∠D)
∠DCE + ∠CDE
= `(1)/(2)(∠"C" + ∠"D")`
= `(1)/(2) xx 180°` = 90°
Thus, in ΔDCE,
∠DEC = 180° - ∠DCE + ∠CDE = 180° - 90°
⇒ ∠DEC = 90°.
APPEARS IN
संबंधित प्रश्न
E is the mid-point of side AB and F is the mid-point of side DC of parallelogram ABCD. Prove that AEFD is a parallelogram.
In the given figure, ABCD is a parallelogram.
Prove that: AB = 2 BC.
Prove that the bisectors of opposite angles of a parallelogram are parallel.
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: QR = QT
Find the perimeter of the parallelogram PQRS.
In the Figure, ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length d of the segment that is perpendicular to `bar("HE")` and `bar("FG")`?
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, it is given that BDEF and FDCE are parallelograms. Can you say that BD = CD? Why or why not?
In the following figure, ABCD and AEFG are two parallelograms. If ∠C = 55º, determine ∠F.
