Question

prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the third side of the triangle.​

Answer 1

$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the third side of the triangle.

$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

${\blue{\sf\underline{Given}}}$

A ∆ABC in which AB=AC an a circle is drawn with AB as diameter, intersecting BC at D.

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BD=CD.

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Join AD

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we have that an angle in a semicircle is a right angle.

∴ ∠ADB=90°. ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀....(i)

Also, BDC being a straight line, we have

⠀⠀⠀⠀⠀⠀⠀∠ADB+∠ADC = 180°

==>90°+∠ADC=180° ⠀⠀[Using (i) ]

==>∠ADC = 90°.

Now, in ∆ADB and ∆ADC , we have

AB=AC ⠀⠀⠀⠀⠀⠀[given]

∠ADB = ∠ADC ⠀[each equal to 90°]

AD=AD ⠀⠀⠀⠀⠀⠀[common]

∴∆ADB ≅ ∆ADC⠀⠀⠀⠀ [by RHS-congruence].

Hence,BD=CD

Answer 2

Answer:

Draw ABC as an ISOSCELes triangle.  On the side  AB (lateral side) mark the mid point  O.  Now as O as the center, draw a circle with radius = OB =OA.  It may not intersect base of all isosceles triangles.  But we choose base BC of our  circle LONG enough so that it will intersect  BC (base) at D.

Now, OB = OA = OD  = radius.

AB = 2 * radius = AC (isosceles triangle)

In triangle  OBD,  anle B = angle D as sides are equal.  Since angle B = angle C, then  angle B = angle C = angle D.

triangles OBD and ABC are similar.  AB || OB,  BD || BC. and  angles are all equal.

as OB = 1/2 AB ,  BD = 1/2 BC.

 Hence the proof is done.