show that the bisector of an angle from vertex of an isosceles triangle bisects the base at right angle
charvig99:
Which chapter and class?
Answers
Answered by
68
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
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
Attachments:
Answered by
10
Step-by-step explanation:
ok it is nice please say thanks and follow me
Attachments:
Similar questions