Advertisements
Advertisements
प्रश्न
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: ∠RTS = 90°
Advertisements
उत्तर
∠PST = ∠TSR
∠QRT = ∠TRS
∠QRS + ∠PSR = 180° ...(adjacent angles of || gm are supplementary)
Multiplying by `(1)/(2)`
`(1)/(2)∠"QRS" + (1)/(2)∠"PSR" = (1)/(2) xx x180°`
∠TSR + ∠TRS = 90°
In ΔSTR,
∠TSR + ∠RTS + ∠TRS = 180°
90° + ∠RTS = 180°
∠RTS = 90°.
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.
In the alongside diagram, ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B.
Prove that:
- AQ = BP
- PQ = CD
- ABPQ is a parallelogram.
Prove that the bisectors of opposite angles of a parallelogram are parallel.
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: RT bisects angle R
In the given figure, MP is the bisector of ∠P and RN is the bisector of ∠R of parallelogram PQRS. Prove that PMRN is a parallelogram.
In a parallelogram ABCD, E is the midpoint of AB and DE bisects angle D. Prove that: BC = BE.
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
Which of the following statement is correct?
In the following figure, it is given that BDEF and FDCE are parallelograms. Can you say that BD = CD? Why or why not?
