. Prove that the circle drawn on anyone of the equal sides of an isosceles triangle as diameter/bisects the base
Answers
step-by-step explanation:
Given:-
In ΔABC,
AB = AC
and a circle with AB as diameter is drawn Which intersects the side BC and D.
To prove:-
D is the mid point of BC
Construction:–
Join AD.
Proof:-
angle 1 = 90°
( angle in semicircule)
angle 1 + angle 2 = 180°
( linear pair)
therefore,
angle 2 = 180° - 90° = 90°
now,
in right ∆ABD and ∆ACD,
AB = AC ( given)
AD = AD ( common )
∴ By the right Angle – Hypotenuse – side criterion of congruence,
we have,
ΔABD ≅ ΔACD [ RHS criterion of congruence]
therefore,
The corresponding parts of the congruent triangle are congruent.
∴ BD = DC [c.p.c.t]
Hence,
D is the mid point of BC.
Thus,
Proved✍️✍️
Draw an isosceles triangle, ΔABC, as an assumption. Here let AB = AC. So that BC is the base which is the unequal side.
Consider the law given below:
If an altitude is drawn from a vertex of an isosceles triangle where equal sides meet each other, then that altitude bisects the third side (base).
According to this law, if an altitude, let it be AD, is drawn from the vertex A, where equal sides AB and AC meet each other,... then AD bisects BC.
As AD is an altitude, AD is the perpendicular bisector of BC such that ∠ADB = ∠ADC = 90°, and D is the midpoint of BC.
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Consider AB. Draw a circle with AB as diameter.
It is true that,
If a circle is drawn with the hypotenuse of a right triangle as the diameter, then the circle passes through the right angle vertex of the triangle.
As ∠ADB = 90°, ΔADB is a right triangle, and if a circle is drawn with AB, the hypotenuse of ΔADB, as diameter, then the circle passes through the right angle vertex of the triangle, i.e., D.
This circle passes through the point D, and D is the midpoint of BC, the base. So the circle bisects BC!!!
Hence proved!
