prove that the circle drawn on any one of the equal sides of an isosce

1.	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» $\huge{\underline{\underline{\sf{\orange{Question:-}}}}}$ 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. $\huge{\underline{\underline{\sf{\orange{Solution:-}}}}}$ ${\blue{\sf\underline{Given}}}$ A ∆ABC in which AB=AC an a circle is drawn with AB as diameter, intersecting BC at D. ${\pink{\sf\underline{To\:Prove}}}$ BD=CD. ${\green{\sf\underline{Construction}}}$ Join AD ${\orange{\sf\underline{Proof}}}$ 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

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»

$\huge{\underline{\underline{\sf{\orange{Question:-}}}}}$

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.

$\huge{\underline{\underline{\sf{\orange{Solution:-}}}}}$

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

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

${\pink{\sf\underline{To\:Prove}}}$

BD=CD.

${\green{\sf\underline{Construction}}}$

Join AD

${\orange{\sf\underline{Proof}}}$

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