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) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB
OR
ABCD 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.
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Answer:
(i) In ΔABC,
It is given that M is the mid-point of AB and MD || BC.
∴ D is the mid-point of AC. [Converse of mid-point theorem]
(ii) As DM || CB and AC is a transversal,
∠MDC + ∠DCB = 180° [Co-interior angles]
∠MDC + 90° = 180°
∠MDC = 90°
∴ MD ⊥ AC
(iii) Join MC
In ΔAMD and ΔCMD,
AD = CD (D is the mid-point of side AC)
∠ADM = ∠CDM (Each 90°)
DM = DM (Common)
∴ ΔAMD ≅ ΔCMD (By SAS congruence rule)
Therefore, AM = CM (By CPCT)
However, we also know that AM = 1/2 AB (M is the mid-point of AB)
∴ CM = AM = 1/2 AB
Step-by-step explanation:
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Verified answer
1.
Given
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To Prove
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Solution
(Refer the diagram below for the given question)
(i) Converse of mid-point theorem states that a line drawn from the mid-point of one side is a triangle, parallel to the other side bisects the third side.
Applying this converse of mid-point theorem, we can say that,
Since,
∴ AD = DC
∴ D is the mid-point of AC
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(ii) Now since DM is parallel to CB, consider AC to be a transversal.
We know that angles on the the same side of the transversal are known as consecutive interior angles and are supplementary.
∴ ∠MDC + ∠DCB = 180°
⇒ ∠MDC + ∠DCB = 180°
⇒ ∠MDC + 90° = 180° [∠DCB is given to be 90° in the question]
⇒ ∠MDC = 180° - 90°
⇒ ∠MDC = 90°
∴ MD ⊥ AC
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(iii) For this sub-part, construct a line joining points M and C.
In ΔADM and ΔMDC,
AD = CD [D is the mid-point of AC]
∠ADM = ∠CDM [90° each]
DM = MD [Common]
∴ ΔADM ≅ ΔCDM by SAS Criteria.
∴ CM = MA by CPCT. - (i)
Since M is the midpoint of AB,
AM = MB = ¹/₂ AB - (ii)
From (i) and (ii)
AM = CM = ¹/₂ AB
Hence, proved.
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2.
Given
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To Prove
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Construction
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Solution
(Refer the diagram below for the given question)
In ΔABC,
According to the mid-point theorem,
A line joining the mid-points of two sides of a triangle is parallel to the third side and is the half of it.
Applying this mid-point theorem,
PQ ll AC
PQ = ¹/₂ AC - (i)
Now, in ΔADC,
Applying this mid-point theorem,
SR ll AC
SR = ¹/₂ AC - (ii)
Taking (i) and (ii),
PQ ll SR
PQ = SR
∴ PQRS is a parallelogram since one pair of opposite sides are equal and parallel.
In ΔSDR and ΔQCR,
DR = RC [R is the midpoint of side CD]
∠SDR = ∠QCR [All angles in a rectangle are 90°]
SD = QC [Opposite sides of rectangle are equal so the half of equal sides are equal as well]
∴ ΔSDR ≅ ΔQCR by SAS Criteria.
∴ SR = QR by CPCT.
SR = PQ since opposite sides of parallelogram are equal and SR = QR since we just proved we can say that,
PQ = QR = SR = PS
∴ Since all sides of the parallelogram are equal, PQRS is a parallelogram.
Hence, proved.
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