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Question
Draw the circumcircle of ΔPMT in which PM = 5.6 cm, ∠P = 60°, ∠M = 70°.
Solution
Step-1: construct the segment PM having length 5.6 cm and using protractor draw lines at the angles ∠P and ∠M which measures 60° and 70° respectively mark the intersection point as T thus ∆PMT is ready
Step-2: construct the perpendicular bisector of line PM by keeping the needle of compass at point P and taking approximately more than half of PM distance in compass draw arc above and below PM
Step-3: keeping the same measurement in compass keep the needle at point M and draw intersecting arcs above and below segment PM
Step-4: join the intersections of arcs to get a line ‘a’ which is the perpendicular bisector of segment PM.
Step-5: Similarly by repeating steps 3,4,5 construct a perpendicular bisector for line TM so instead of P substitute T and repeat steps 3,4,5 we will get a line ‘b’ perpendicular bisector of segment TM
Step-6: keep the needle of compass at point of intersection of the line a and b and from there take distance till any vertex of triangle PTM and construct the circle
The circle is required circumcircle to ∆PTM
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∴ `square/square = square/square` .......... (I) theorem of angle bisector.
In △PMR, ray MY is bisector of ∠PMQ.
∴ `square/square = square/square` .......... (II) theorem of angle bisector.
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solution:
In ∆PMQ,
Ray MX is the bisector of ∠PMQ.
∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]
Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.
∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]
But `("MP")/("MQ") = ("MP")/("MR")` .............(III) [As M is the midpoint of QR.]
Hence MQ = MR
∴ `("PX")/square = square/("YR")` .............[From (I), (II) and (III)]
∴ XY || QR .............[Converse of basic proportionality theorem]
In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:
Proof :
In ΔABC, ray BD bisects ∠B.
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∴ `(square)/("EB") = ("AD")/("DC")` ...(II) (`square`)
∴ `("AB")/square = square/("EB")` ...[from (I) and (II)]
