Advertisements
Advertisements
Question
In the below Fig, ABCD and PQRC are rectangles and Q is the mid-point of Prove thaT
i) DP = PC (ii) PR = `1/2` AC

Advertisements
Solution

(i)In DADC, Q is the midpoint of AC such that
PQ || AD
∴ P is the midpoint of DC
⇒ DP = DC [Using converse of midpoint theorem]
(ii)Similarly, R is the midpoint of BC
∴ PR = `1/2` BD [Diagonal of rectangle are equal \BD = AC ]
PR = `1/2` AC
APPEARS IN
RELATED QUESTIONS
ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively
intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of
ΔABC
In the given figure, `square`PQRS and `square`MNRL are rectangles. If point M is the midpoint of side PR then prove that,
- SL = LR
- LN = `1/2`SQ

The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:
PR = `[1]/[2]` ( AB + CD)

In triangle ABC ; D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F.
Prove that BDEF is a parallelogram. If AB = 16 cm, AC = 12 cm and BC = 18 cm,
find the perimeter of the parallelogram BDEF.
In parallelogram PQRS, L is mid-point of side SR and SN is drawn parallel to LQ which meets RQ produced at N and cuts side PQ at M. Prove that M is the mid-point of PQ.
ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: 2EF = BD.
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: The line drawn through G and parallel to FE and bisects DA.
In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the mid-points of AB and BC, determine the length of DE.
D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.
