Question

show that the bisector of an angle from vertex of an isosceles triangle bisects the base at right angle

Answer 1

In the ΔABD and ΔADC
1)AB=AC(GIVEN)
2)BAD=CAD(GIVEN)
3)AD (COMMON)
SO ΔABD≡ΔACD(S-A-S)
  ⇒ADB=ADC(cpctc)
⇒BD=CD(cpctc)  ⇒ AD bisects BC

we know,
ADB+ABC=180°(supplementary angles)
⇒2X ADB=180°
⇒ADB=90°

⇒AD is perpendicular to BC

Answer 2

Step-by-step explanation:

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