Question

7. ABC is an isosceles triangle with AB = AC and
AD is the altitude of the triangle. Prove that AD is
also a median of the triangle.​

Answer 1

Answer:  Step-by-step explanation: Given: ABC is an isosceles triangle such that  AB=AC AD  is the altitude  To Prove :ar(ABD)=ar(ACD)  Proof: Consider triangle ABD and triangle ACD                      <ADB=<ADC (altitude is also a perpendicular bisector according to the       triangle property)              AD=AD(common)              AB=AC(given)            By RHS congruency                         ΔABD≅ΔACD We know that the triangle which are congruent have the same area ∴ ar(ABD)=ar(ACD) AD is the median of the ΔABC

Answer 2

$\huge\mathfrak{Answer}$

Here Given that

$\triangle$ABC is isosceles with

AB = AC and AB$\perpendicular$BC

We have to prove that AD is also the median of the triangle.

In$\triangle$s ABD and ACD,

$\angle$ADB = $\angle$ADC [each = 90° (given)]

AB = AC (given)

AD = AD (common)

$\triangle$ABD $\cong$ $\triangle$ACB (By RHS congruency rule)

=> BD = CD ( c.p.c.t)

=> D is the mid point of BC i.e, altitude AD divides BC in two equal parts.

Thus altitude AD is also a median of the triangle ABC.

#Answerwithquality

#BAL