In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB=AC
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Answer:
In ∆ABD and ∆DAC
BD = DC (Since AD Perpendicular to BC)
✓ BDA =✓ CDA = 90° (AD perpendicular which makes 90°)
AD=AD (common)
Therefore, ∆ABD congurent∆DAC
AB=AC (by (CPCT)
Hence, proved
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Answer:
given, triangle ABC
and AD is perpendicular bisector of BC
SO, angle ADB = angle ADC = 90°
and BD = DC
by using phythagorean theorem,
AB^2 = AD^2 + BD^2 ( in ADB) - eq.1
AC^2 = AD^2+ DC^2 ( in ADC)
{ BD = DC ,on squaring both side
BD^2 = DC^2 }
AC^2 = AD^2+ BD^2 ( in ADC) - eq. 2
On equating, equation 1 and 2
AB^2 = AC^2
taking square root on both sides,
AB = AC
HENCE, PROVED
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