In ΔABC, P is mid-point to AB [∵ Given] Q is mid-point of BC [∵ Given] Hence, PQ ∥ AC and PQ = (1/2)AC ... (1) [∵ Mid-point Theorem] Similarly, in ΔACD, S is mid-point of AD [∵ Given] R is mid-point of CD [∵ Given] Hence, SR ∥ AC and SR = (1/2) AC ...(2) [∵ Mid-Point Theorem] From (1) and (2), we haRead more
In ΔABC,
P is mid-point to AB [∵ Given]
Q is mid-point of BC [∵ Given]
Hence, PQ ∥ AC and PQ = (1/2)AC … (1) [∵ Mid-point Theorem]
Similarly, in ΔACD,
S is mid-point of AD [∵ Given]
R is mid-point of CD [∵ Given]
Hence, SR ∥ AC and SR = (1/2) AC …(2) [∵ Mid-Point Theorem]
From (1) and (2), we have
PQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]
Hence, PQRS is a parallelogram.
Similarly, in ΔBCD,
Q is mid-point of BC [∵ Given]
R is mid-point of CD [∵ Given]
Hence, QR = (1/2) BD …(5) [∵ Mid-point theorem]
Given that: AC = BD …(6) [∵ Diagonals of a rectangle are equal]
From (1), (5) and (6), we have
PQ = QR
A parallelogram whose adjacent sides are equal, is a rhombus. Hence, PQRS is a rhombus.
In ΔACD, S is mid-point to DA [∵ Given] R is mid-point of DC [∵ Given] Hence, SR ∥ AC and SR = (1/2)AC ... (1) [∵ Mid-point Theorem] In ΔACD, P is mid-point of AB [∵ Given] Q is mid-point of BC [∵ Given] Hence, PQ ∥ AC and PQ = (1/2) AC ...(2) [∵ Mid-Point Theorem] From (1) and (2), we have PQ ∥ SRRead more
In ΔACD,
S is mid-point to DA [∵ Given]
R is mid-point of DC [∵ Given]
Hence, SR ∥ AC and SR = (1/2)AC … (1) [∵ Mid-point Theorem]
In ΔACD,
P is mid-point of AB [∵ Given]
Q is mid-point of BC [∵ Given]
Hence, PQ ∥ AC and PQ = (1/2) AC …(2) [∵ Mid-Point Theorem]
From (1) and (2), we have
PQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]
In quadrilateral PQRS,
PQ ∥ SR and PQ = SR [∵ From (3) and (4)]
Hence, PQRS is a parallelogram and diagonals of parallelogram bisect each other.
Therefore, SQ and PR bisects each other.
(i) In ΔABC, M is mid-point to AB [∵ Given] and DM ∥ BC [∵ Given] Hence, D is mid-point of AC [∵ Converse of Mid-point Theorem] (ii) ∠ADM = ∠ACB [∵ Corresponding angles] ⇒ ∠ADM = 90° [∵ ∠ACB 90°] Hence, MD ⊥ AC (iii) In ΔAMD = ΔCMD, AD = DC [∵ Proved above] ∠ADM = ∠CDM [∵ Each 90°] DM = DM [∵ CommonRead more
(i) In ΔABC,
M is mid-point to AB [∵ Given]
and DM ∥ BC [∵ Given]
Hence, D is mid-point of AC [∵ Converse of Mid-point Theorem]
(ii) ∠ADM = ∠ACB [∵ Corresponding angles]
⇒ ∠ADM = 90° [∵ ∠ACB 90°]
Hence, MD ⊥ AC
(iii) In ΔAMD = ΔCMD,
AD = DC [∵ Proved above]
∠ADM = ∠CDM [∵ Each 90°]
DM = DM [∵ Common]
Hence, ΔAMD ΔCMD [∵ SAS Congruency rule]
AM = CM [∵ CPCT]
But AM = (1/2) AB [∵ Given]
Therefore, CM = AM = (1/2) AB
ABC is a triangle. Locate a point in the interior of Triangle ABC which is equidistant from all the vertices of Triangle ABC.
Draw perpendicular bisectors of AB, BC, and AC, which intersects each other at G. Point G is equidistant from the three vertices of ΔABC.
Draw perpendicular bisectors of AB, BC, and AC, which intersects each other at G.
See lessPoint G is equidistant from the three vertices of ΔABC.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Draw the bisectors of ∠A, ∠B and ∠C, which intersect each other at O. Point O is equidistant from the three sides of ΔABC ie. OP = OQ = OR.
Draw the bisectors of ∠A, ∠B and ∠C, which intersect each other at O. Point O is equidistant from the three sides of ΔABC ie. OP = OQ = OR.
See lessABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
In ΔABC, P is mid-point to AB [∵ Given] Q is mid-point of BC [∵ Given] Hence, PQ ∥ AC and PQ = (1/2)AC ... (1) [∵ Mid-point Theorem] Similarly, in ΔACD, S is mid-point of AD [∵ Given] R is mid-point of CD [∵ Given] Hence, SR ∥ AC and SR = (1/2) AC ...(2) [∵ Mid-Point Theorem] From (1) and (2), we haRead more
In ΔABC,
P is mid-point to AB [∵ Given]
Q is mid-point of BC [∵ Given]
Hence, PQ ∥ AC and PQ = (1/2)AC … (1) [∵ Mid-point Theorem]
Similarly, in ΔACD,
S is mid-point of AD [∵ Given]
R is mid-point of CD [∵ Given]
Hence, SR ∥ AC and SR = (1/2) AC …(2) [∵ Mid-Point Theorem]
From (1) and (2), we have
PQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]
Hence, PQRS is a parallelogram.
Similarly, in ΔBCD,
See lessQ is mid-point of BC [∵ Given]
R is mid-point of CD [∵ Given]
Hence, QR = (1/2) BD …(5) [∵ Mid-point theorem]
Given that: AC = BD …(6) [∵ Diagonals of a rectangle are equal]
From (1), (5) and (6), we have
PQ = QR
A parallelogram whose adjacent sides are equal, is a rhombus. Hence, PQRS is a rhombus.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
In ΔACD, S is mid-point to DA [∵ Given] R is mid-point of DC [∵ Given] Hence, SR ∥ AC and SR = (1/2)AC ... (1) [∵ Mid-point Theorem] In ΔACD, P is mid-point of AB [∵ Given] Q is mid-point of BC [∵ Given] Hence, PQ ∥ AC and PQ = (1/2) AC ...(2) [∵ Mid-Point Theorem] From (1) and (2), we have PQ ∥ SRRead more
In ΔACD,
S is mid-point to DA [∵ Given]
R is mid-point of DC [∵ Given]
Hence, SR ∥ AC and SR = (1/2)AC … (1) [∵ Mid-point Theorem]
In ΔACD,
P is mid-point of AB [∵ Given]
Q is mid-point of BC [∵ Given]
Hence, PQ ∥ AC and PQ = (1/2) AC …(2) [∵ Mid-Point Theorem]
From (1) and (2), we have
See lessPQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]
In quadrilateral PQRS,
PQ ∥ SR and PQ = SR [∵ From (3) and (4)]
Hence, PQRS is a parallelogram and diagonals of parallelogram bisect each other.
Therefore, SQ and PR bisects each other.
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) In ΔABC, M is mid-point to AB [∵ Given] and DM ∥ BC [∵ Given] Hence, D is mid-point of AC [∵ Converse of Mid-point Theorem] (ii) ∠ADM = ∠ACB [∵ Corresponding angles] ⇒ ∠ADM = 90° [∵ ∠ACB 90°] Hence, MD ⊥ AC (iii) In ΔAMD = ΔCMD, AD = DC [∵ Proved above] ∠ADM = ∠CDM [∵ Each 90°] DM = DM [∵ CommonRead more
(i) In ΔABC,
M is mid-point to AB [∵ Given]
and DM ∥ BC [∵ Given]
Hence, D is mid-point of AC [∵ Converse of Mid-point Theorem]
(ii) ∠ADM = ∠ACB [∵ Corresponding angles]
⇒ ∠ADM = 90° [∵ ∠ACB 90°]
Hence, MD ⊥ AC
(iii) In ΔAMD = ΔCMD,
See lessAD = DC [∵ Proved above]
∠ADM = ∠CDM [∵ Each 90°]
DM = DM [∵ Common]
Hence, ΔAMD ΔCMD [∵ SAS Congruency rule]
AM = CM [∵ CPCT]
But AM = (1/2) AB [∵ Given]
Therefore, CM = AM = (1/2) AB