Question

ABCD is the triangle right angled at c a line through the midpoint of hypotenuse AC and parallel to BC intersects AC And M show that
(i) M is the midpoint of Ac,
(ii) MP⊥AC,
(iii)CP = AP = ½AB​

Answer 1

Answer:

In △ABC,

(1) M is the mid point of AB and

MD∥BC

D is the mid point of AC

(2) As MD∥BC &

AC is transversal

∠MDC+∠BCD=180

∘

∠MDC+90

∘

=180

∘

⇒∠MDC=90

∘

⇒MD⊥AC

(3) In △AMD and △CMD

AD=CD

∠ADM=∠CDM

DM=DM

△AMD≅△CMD

AM=CM⇒AM=

2

1

AB⇒CM=AM=

2

1

AB

Answer 2

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$\huge\underline\bold\orange{Question}$

ABCD is the triangle right angled at c a line through the midpoint of hypotenuse AC and parallel to BC intersects AC And M show that

(i) M is the midpoint of Ac,

(ii) MP⊥AC,

(iii)CP = AP = ½AB

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$\huge\underline\bold\orange{Solution}$

In ∆ACB, we have

⠀⠀⠀P is the midpoint of AB and PM || BC.

∴ M is the midpoint of AC.

⠀⠀⠀[by Converse of midpoint theorem]

Now, PM || BC.

∴ ∠PMC + ∠BCM = 180° (co-interior ∠s)

⇒ ∠PMC + 90° = 180° ⇒ ∠PMC = 90°.

Thus, MP⊥AC.

Join PC.

In ∆PMA and PMC, we have:

⠀⠀⠀MA = MC [∴M is the midpoint of AC]

⠀⠀⠀∠PMA = ∠PMC (each equal to 90°)

⠀⠀⠀PM = PM (common)

∴ ∆PMA ≅ ∆PMC (SAS-criterion).

And so,AP = CP (c.p.c.t.).

Now, P is the midpoint of AB.

$\therefore \: cp = ap = \frac{1}{2} ab.$

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$\sf{\underline{\underline{\pink{Additional\:Information-}}}}$

✯︎The converse of the midpoint theorem states that ” if a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, it bisects the third side”.

✯︎Formula of midpoint

• Midpoint = [(x1 + x2)/2, (y1 + y2)/2]

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