In $ ∆\mathrm{ABC}$ $ \mathrm{AD}$ is the perpendicular bisector of $ \mathrm{BC}$. Show that $ ∆\mathrm{ABC}$ is an isosceles triangle in which $ \mathrm{AB} = \mathrm{AC}$.

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Solution

Prove the required statement:
Given: $AD is perpendicular bisector of BC$
$In ∆ ADB and ∆ ADC AD = AD (Common side) ∠ ADB = ∠ ADC (90 ° angle) BD = DC (AD is bisector of BC) ∴ ∆ ADB ≅ ∆ ADC (by SAS congruency) \Rightarrow AB = AC (Corresponding sides of congruent triangles)$
Hence, proved that $AB = AC$ .

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In $ ∆\mathrm{ABC}$ $ \mathrm{AD}$ is the perpendicular bisector of $ \mathrm{BC}$. Show that $ ∆\mathrm{ABC}$ is an isosceles triangle in which $ \mathrm{AB} = \mathrm{AC}$.

Prove the required statement:Given: ADisperpendicularbisectorofBCIn∆ADBand∆ADCAD=AD(Commonside)∠ADB=∠ADC(90°angle)BD=DC(ADisbisectorofBC)∴∆ADB≅∆ADC(bySAScongruency)⇒AB=AC(Correspondingsidesofcongruenttriangles)Hence, proved that AB=AC.