Question

Prove that triangle ABC is isosceles if median AD is perpendicular to BC.
(ii) If in the same triangle ABC taken in part (i) the vertex angle A = 90° , then find the value of other two base angles.

Answer 1

Step-by-step explanation:

now ABC has two more triangels in it ADB and ADC

so in ADC and ADB

AD = AD ( as a common side )

<ADC = < ADB=90 ( as AD is perpendicular to BC)

<BAD = < CAD (as AD is perpendicular to BC so it will also be angle bisector of <A)

which means ADC is congruent to triangle ADB by ( ASA)

so AC = AB by (CPCT)

which means two sides of triangle ABC is equal

so this makes it an isoceles triangle

1) now if angle A is 90 degrees in triangle ABC

and it is an isoceles triangle so it will have its rest of two angles equal

let those angles be = x

so

90 + x+ x = 180

2x = 180 - 90

x = 90 / 2 = 45 deg

sorry i am unable to answer rest of the question